{"id":600986,"date":"2022-05-11T10:34:31","date_gmt":"2022-05-11T10:34:31","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=600986"},"modified":"2022-05-13T04:52:44","modified_gmt":"2022-05-13T04:52:44","slug":"ml-aggarwal-solutions-for-class-9-maths-chapter-4-factorization","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-9-maths-chapter-4-factorization\/","title":{"rendered":"ML Aggarwal Solutions for Class 9 Maths Chapter 4- Factorization"},"content":{"rendered":"\n

Class 9: Maths Chapter 4 solutions. Complete Class 9 Maths Chapter 4 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 9 Maths Chapter 4- Factorization<\/h2>\n\n\n\n

ML Aggarwal 9th Maths Chapter 4, Class 9 Maths Chapter 4 solutions<\/p>\n\n\n\n

Exercise 4.1<\/h4>\n\n\n\n

Factorise the following (1 to 9):<\/strong><\/p>\n\n\n\n

1. (i) 8xy3<\/sup> + 12x2<\/sup>y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

8xy3<\/sup> + 12x2<\/sup>y2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 4xy2<\/sup> (2y + 3x)<\/p>\n\n\n\n

Therefore, HCF of 8xy3<\/sup> and 12x2<\/sup>y2<\/sup> is 4xy2<\/sup>.<\/p>\n\n\n\n

(ii) 15 ax3<\/sup> \u2013 9ax2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

15 ax3<\/sup> \u2013 9ax2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 3ax2<\/sup> (5x \u2013 3)<\/p>\n\n\n\n

Therefore, HCF of 15 ax3<\/sup> and 9ax2<\/sup> is 3ax2<\/sup>.<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) 21py2<\/sup> \u2013 56py<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

21py2<\/sup> \u2013 56py<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 7py (3y \u2013 8)<\/p>\n\n\n\n

Therefore, HCF of 21py2<\/sup> and 56py is 7py.<\/p>\n\n\n\n

(ii) 4x3<\/sup> \u2013 6x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4x3<\/sup> \u2013 6x2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 2x2<\/sup> (2x \u2013 3)<\/p>\n\n\n\n

Therefore, HCF of 4x3<\/sup> and 6x2<\/sup> is 2x2<\/sup>.<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(i) 2\u03c0r2<\/sup> \u2013 4\u03c0r<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2\u03c0r2<\/sup> \u2013 4\u03c0r<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 2\u03c0r (r \u2013 2)<\/p>\n\n\n\n

Therefore, HCF of 2\u03c0r2<\/sup> and 4\u03c0r is 2\u03c0r.<\/p>\n\n\n\n

(ii) 18m + 16n<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

18m + 16n<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 2 (9m \u2013 8n)<\/p>\n\n\n\n

Therefore, HCF of 18m and 16n is 2.<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) 25abc2<\/sup> \u2013 15a2<\/sup>b2<\/sup>c<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

25abc2<\/sup> \u2013 15a2<\/sup>b2<\/sup>c<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 5abc (5c \u2013 3ab)<\/p>\n\n\n\n

Therefore, HCF of 25abc2<\/sup> and 15a2<\/sup>b2<\/sup>c is 5abc.<\/p>\n\n\n\n

(ii) 28p2<\/sup>q2<\/sup>r \u2013 42pq2<\/sup>r2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

28p2<\/sup>q2<\/sup>r \u2013 42pq2<\/sup>r2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 14pq2<\/sup>r (2p \u2013 3r)<\/p>\n\n\n\n

Therefore, HCF of 28p2<\/sup>q2<\/sup>r and 42pq2<\/sup>r2<\/sup> is 14pq2<\/sup>r.<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

(i) 8x3<\/sup> \u2013 6x2<\/sup> + 10x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

8x3<\/sup> \u2013 6x2<\/sup> + 10x<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 2x(4x2<\/sup> \u2013 3x + 5)<\/p>\n\n\n\n

Therefore, HCF of 8x3<\/sup>, 6x2<\/sup> and 10x is 2x.<\/p>\n\n\n\n

(ii) 14mn + 22m \u2013 62p<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

14mn + 22m \u2013 62p<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 2 (7mn + 11m \u2013 31p)<\/p>\n\n\n\n

Therefore, HCF of 14mn, 22m and 62p is 2.<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) 18p2<\/sup>q2<\/sup> \u2013 24pq2<\/sup> + 30p2<\/sup>q<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

18p2<\/sup>q2<\/sup> \u2013 24pq2<\/sup> + 30p2<\/sup>q<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 6pq (3pq \u2013 4q + 5p)<\/p>\n\n\n\n

Therefore, HCF of 18p2<\/sup>q2<\/sup>, 24pq2<\/sup> and 30p2<\/sup>q is 6pq.<\/p>\n\n\n\n

(ii) 27a3<\/sup>b3<\/sup> \u2013 18a2<\/sup>b3<\/sup> + 75a3<\/sup>b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

27a3<\/sup>b3<\/sup> \u2013 18a2<\/sup>b3<\/sup> + 75a3<\/sup>b2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 3a2<\/sup>b2<\/sup> (9a \u2013 6b + 25a)<\/p>\n\n\n\n

Therefore, HCF of 27a3<\/sup>b3<\/sup>, 18a2<\/sup>b3<\/sup> and 75a3<\/sup>b2<\/sup> is 3a2<\/sup>b2<\/sup>.<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) 15a (2p \u2013 3q) \u2013 10b (2p \u2013 3q)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

15a (2p \u2013 3q) \u2013 10b (2p \u2013 3q)<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 5(2p \u2013 3q) [3a \u2013 2b]<\/p>\n\n\n\n

Therefore, HCF of 15a (2p \u2013 3q) and 10b (2p \u2013 3q) is 5(2p \u2013 3q).<\/p>\n\n\n\n

(ii) 3a(x2<\/sup> + y2<\/sup>) + 6b (x2<\/sup> + y2<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

3a(x2<\/sup> + y2<\/sup>) + 6b (x2<\/sup> + y2<\/sup>)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

Then, 3(x2<\/sup> + y2<\/sup>) (a + 2b)<\/p>\n\n\n\n

Therefore, HCF of 3a(x2<\/sup> + y2<\/sup>) and 6b (x2<\/sup> + y2<\/sup>) is 3(x2<\/sup> + y2<\/sup>).<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) 6(x + 2y)3<\/sup> + 8(x + 2y)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

6(x + 2y)3<\/sup> + 8(x + 2y)2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

Then, 2(x + 2y)2<\/sup> [3(x + 2y) + 4]<\/p>\n\n\n\n

Therefore, HCF of 6(x + 2y)3<\/sup> and 8(x + 2y)2<\/sup> is 2(x + 2y)2<\/sup>.<\/p>\n\n\n\n

(ii) 14(a \u2013 3b)3<\/sup> \u2013 21p(a \u2013 3b)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

14(a \u2013 3b)3<\/sup> \u2013 21p(a \u2013 3b)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

Then, 7(a \u2013 3b) [2(a \u2013 3b)2<\/sup> \u2013 3p]<\/p>\n\n\n\n

Therefore, HCF of 14(a \u2013 3b)3<\/sup> and 21p(a \u2013 3b) is 7(a \u2013 3b).<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) 10a(2p + q)3<\/sup> \u2013 15b (2p + q)2<\/sup> + 35 (2p + q)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

10a(2p + q)3<\/sup> \u2013 15b (2p + q)2<\/sup> + 35 (2p + q)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

Then, 5(2p + q) [2a (2p + q)2<\/sup> \u2013 3b (2p + q) + 7]<\/p>\n\n\n\n

Therefore, HCF of 10a(2p + q)3<\/sup>, 15b (2p + q)2<\/sup> and 35 (2p + q) is 5(2p + q).<\/p>\n\n\n\n

(ii) x(x2<\/sup> + y2<\/sup> \u2013 z2<\/sup>) + y(-x2<\/sup> \u2013 y2<\/sup> + z2<\/sup>) \u2013 z (x2<\/sup> + y \u2013 z2<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x(x2<\/sup> + y2<\/sup> \u2013 z2<\/sup>) + y(-x2<\/sup> \u2013 y2<\/sup> + z2<\/sup>) \u2013 z (x2<\/sup> + y \u2013 z2<\/sup>)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

Then, (x2<\/sup> + y2<\/sup> \u2013 z2<\/sup>) [x \u2013 y \u2013 z]<\/p>\n\n\n\n

Therefore, HCF of x(x2<\/sup> + y2<\/sup> \u2013 z2<\/sup>), y(-x2<\/sup> \u2013 y2<\/sup> + z2<\/sup>) and z (x2<\/sup> + y \u2013 z2<\/sup>) is (x2<\/sup> + y2<\/sup> \u2013 z2<\/sup>)<\/p>\n\n\n\n

Exercise 4.2<\/h4>\n\n\n\n

Factorise the following (1 to 13):<\/strong><\/p>\n\n\n\n

1.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + xy \u2013 x \u2013 y<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> + xy \u2013 x \u2013 y<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(x + y) \u2013 1(x + y)<\/p>\n\n\n\n

(x + y) (x \u2013 1)<\/p>\n\n\n\n

(ii) y2<\/sup> \u2013 yz \u2013 5y + 5z<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

y2<\/sup> \u2013 yz \u2013 5y + 5z<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

y(y \u2013 z) \u2013 5(y \u2013 z)<\/p>\n\n\n\n

(y \u2013 z) (y \u2013 5)<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) 5xy + 7y \u2013 5y2<\/sup> \u2013 7x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5xy \u2013 7x \u2013 5y2<\/sup> + 7y<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(5y \u2013 7) \u2013 y(5y \u2013 7)<\/p>\n\n\n\n

(5y \u2013 7) (x \u2013 y)<\/p>\n\n\n\n

(ii) 5p2<\/sup> \u2013 8pq \u2013 10p + 16q<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5p2<\/sup> \u2013 8pq \u2013 10p + 16q<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

p(5p \u2013 8q) \u2013 2(5p \u2013 8q)<\/p>\n\n\n\n

(5p \u2013 8q) (p \u2013 2)<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(i) a2<\/sup>b \u2013 ab2<\/sup> + 3a \u2013 3b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a2<\/sup>b \u2013 ab2<\/sup> + 3a \u2013 3b<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

ab(a \u2013 b) + 3(a \u2013 b)<\/p>\n\n\n\n

(a \u2013 b) (ab + 3)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 3x2<\/sup> + x \u2013 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 3x2<\/sup> + x \u2013 3<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x2<\/sup> (x \u2013 3) + 1(x \u2013 3)<\/p>\n\n\n\n

(x \u2013 3) (x2<\/sup> + 1)<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) 6xy2<\/sup> \u2013 3xy \u2013 10y + 5<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

6xy2<\/sup> \u2013 3xy \u2013 10y + 5<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

3xy(2y \u2013 1) \u2013 5(2y \u2013 1)<\/p>\n\n\n\n

(2y \u2013 1) (3xy \u2013 5)<\/p>\n\n\n\n

(ii) 3ax \u2013 6ay \u2013 8by + 4bx<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

3ax \u2013 6ay \u2013 8by + 4bx<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

3a(x \u2013 2y) + 4b (x \u2013 2y)<\/p>\n\n\n\n

(x \u2013 2y) (3a + 4b)<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

(i) 1 \u2013 a \u2013 b + ab<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

1 \u2013 a \u2013 b + ab<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

1(1 \u2013 a) \u2013 b(1 \u2013 a)<\/p>\n\n\n\n

(1 \u2013 a) (1 \u2013 b)<\/p>\n\n\n\n

(ii) a(a \u2013 2b \u2013 c) + 2bc<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a(a \u2013 2b \u2013 c) + 2bc<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

a2<\/sup> \u2013 2ab \u2013 ac + 2bc<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

a(a \u2013 2b) \u2013 c(a + 2b)<\/p>\n\n\n\n

(a \u2013 2b) (a \u2013 c)<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + xy (1 + y) + y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> + xy (1 + y) + y3<\/sup><\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

x2<\/sup> + xy + xy2<\/sup> + y3<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(x + y) + y2<\/sup>(x + y)<\/p>\n\n\n\n

(x + y) (x + y2<\/sup>)<\/p>\n\n\n\n

(ii) y2<\/sup> \u2013 xy (1 \u2013 x) \u2013 x3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

y2<\/sup> \u2013 xy (1 \u2013 x) \u2013 x3<\/sup><\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

y2<\/sup> \u2013 xy + x2<\/sup>y \u2013 x3<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

y(y \u2013 x) + x2<\/sup> (y \u2013 x)<\/p>\n\n\n\n

(y \u2013 x) (y + x2<\/sup>)<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) ab2<\/sup> + (a \u2013 1)b \u2013 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

ab2<\/sup> + (a \u2013 1)b \u2013 1<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

ab2<\/sup> + ab \u2013 b \u2013 1<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

ab(b + 1) \u2013 1(b + 1)<\/p>\n\n\n\n

(b + 1) (ab \u2013 1)<\/p>\n\n\n\n

(ii) 2a \u2013 4b \u2013 xa + 2bx<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2a \u2013 4b \u2013 xa + 2bx<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2(a \u2013 2b) \u2013 x(a \u2013 2b)<\/p>\n\n\n\n

(a \u2013 2b) (2 \u2013 x)<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) 5ph \u2013 10qk + 2rph \u2013 4qrk<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5ph \u2013 10qk + 2rph \u2013 4qrk<\/p>\n\n\n\n

Re-arranging the given question we get,<\/p>\n\n\n\n

5ph + 2rph \u2013 10qk \u2013 4qrk<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

ph(5 + 2r) \u2013 2qk(5 + 2r)<\/p>\n\n\n\n

(5 + 2r) (ph \u2013 2qk)<\/p>\n\n\n\n

(ii) x2<\/sup> \u2013 x(a + 2b) + 2ab<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 x(a + 2b) + 2ab<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

x2<\/sup> \u2013 xa \u2013 2xb + 2ab<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(x \u2013 a) \u2013 2b(x \u2013 a)<\/p>\n\n\n\n

(x \u2013 a) (x \u2013 2b)<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) ab(x2<\/sup> + y2<\/sup>) \u2013 xy(a2<\/sup> + b2<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

ab(x2<\/sup> + y2<\/sup>) \u2013 xy(a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

abx2<\/sup> + aby2<\/sup> \u2013 xya2<\/sup> \u2013 xyb2<\/sup><\/p>\n\n\n\n

Re-arranging the above we get,<\/p>\n\n\n\n

abx2<\/sup> \u2013 xyb2<\/sup> + aby2<\/sup> \u2013 xya2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

bx(ax \u2013 by) + ay(by \u2013 ax)<\/p>\n\n\n\n

bx(ax \u2013 by) \u2013 ay (ax \u2013 by)<\/p>\n\n\n\n

(ax \u2013 by) (bx \u2013 ay)<\/p>\n\n\n\n

(ii) (ax + by)2<\/sup> + (bx \u2013 ay)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By expanding the give question, we get,<\/p>\n\n\n\n

(ax)2<\/sup> + (by)2<\/sup> + 2axby + (bx)2<\/sup> + (ay)2<\/sup> \u2013 2bxay<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + b2<\/sup>y2<\/sup> + b2<\/sup>x2<\/sup> + a2<\/sup>y2<\/sup><\/p>\n\n\n\n

Re-arranging the above we get,<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + a2<\/sup>y2<\/sup> + b2<\/sup>y2<\/sup> + b2<\/sup>x2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

a2<\/sup> (x2<\/sup> + y2<\/sup>) + b2<\/sup> (x2<\/sup> + y2<\/sup>)<\/p>\n\n\n\n

(x2<\/sup> + y2<\/sup>) (a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

(i) a3<\/sup> + ab(1 \u2013 2a) \u2013 2b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> + ab(1 \u2013 2a) \u2013 2b2<\/sup><\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

a3<\/sup> + ab \u2013 2a2<\/sup>b \u2013 2b2<\/sup><\/p>\n\n\n\n

Re-arranging the above we get,<\/p>\n\n\n\n

a3<\/sup> \u2013 2a2<\/sup>b + ab \u2013 2b2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

a2<\/sup>(a \u2013 2b) + b(a \u2013 2b)<\/p>\n\n\n\n

(a \u2013 2b) (a2<\/sup> + b)<\/p>\n\n\n\n

(ii) 3x2<\/sup>y \u2013 3xy + 12x \u2013 12<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

3x2<\/sup>y \u2013 3xy + 12x \u2013 12<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

3xy(x \u2013 1) + 12(x \u2013 1)<\/p>\n\n\n\n

(x \u2013 1) (3xy + 12)<\/p>\n\n\n\n

11. a2<\/sup>b + ab2<\/sup> \u2013abc \u2013 b2<\/sup>c + axy + bxy<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a2<\/sup>b + ab2<\/sup> \u2013abc \u2013 b2<\/sup>c + axy + bxy<\/p>\n\n\n\n

Re-arranging the above we get,<\/p>\n\n\n\n

a2<\/sup>b \u2013 abc + axy + ab2<\/sup> \u2013 b2<\/sup>c + bxy<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

a(ab \u2013 bc + xy) + b(ab \u2013 bc + xy)<\/p>\n\n\n\n

(a + b) (ab \u2013 bc + xy)<\/p>\n\n\n\n

12. ax2<\/sup> \u2013 bx2<\/sup> + ay2<\/sup> \u2013 by2<\/sup> + az2<\/sup> \u2013 bz2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

ax2<\/sup> \u2013 bx2<\/sup> + ay2<\/sup> \u2013 by2<\/sup> + az2<\/sup> \u2013 bz2<\/sup><\/p>\n\n\n\n

Re-arranging the above we get,<\/p>\n\n\n\n

ax2<\/sup> + ay2<\/sup> + az2<\/sup> \u2013 bx2<\/sup> \u2013 by2<\/sup> \u2013 bz2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

a(x2<\/sup> + y2<\/sup> + z2<\/sup>) \u2013 b(x2<\/sup> + y2<\/sup> + z2<\/sup>)<\/p>\n\n\n\n

(x2<\/sup> + y2<\/sup> + z2<\/sup>) (a \u2013 b)<\/p>\n\n\n\n

13. x \u2013 1 \u2013 (x \u2013 1)2<\/sup> + ax \u2013 a<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x \u2013 1 \u2013 (x \u2013 1)2<\/sup> + ax \u2013 a<\/p>\n\n\n\n

By expanding the above we get,<\/p>\n\n\n\n

X \u2013 1 \u2013 (x2<\/sup> + 1 \u2013 2x) + ax \u2013 a<\/p>\n\n\n\n

x \u2013 1 \u2013 x2<\/sup> -1 + 2x + ax \u2013 a<\/p>\n\n\n\n

2x \u2013 x2<\/sup> + ax \u2013 2 + x \u2013 a<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(2 \u2013 x + a) \u2013 1(2 \u2013 x + a)<\/p>\n\n\n\n

(2 \u2013 x + a) (x \u2013 1)<\/p>\n\n\n\n

Exercise 4.3<\/h4>\n\n\n\n

Factorise the following (1 to 17):<\/strong><\/p>\n\n\n\n

1. 4x2<\/sup> \u2013 25y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, (2x)2<\/sup> \u2013 (5y)2<\/sup><\/p>\n\n\n\n

Then, (2x + y) (2x \u2013 5y)<\/p>\n\n\n\n

(ii) 9x2<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, (3x)2<\/sup> \u2013 12<\/sup><\/p>\n\n\n\n

Then, (3x + 1) (3x \u2013 1)<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) 150 \u2013 6a2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

150 \u2013 6a2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

6(25 \u2013 a2<\/sup>)<\/p>\n\n\n\n

6(52<\/sup> \u2013 a2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, 6(5 + a) (5 \u2013 a)<\/p>\n\n\n\n

(ii) 32x2<\/sup> \u2013 18y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

32x2<\/sup> \u2013 18y2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2(16x2<\/sup> \u2013 9y2<\/sup>)<\/p>\n\n\n\n

2((4x)2<\/sup> \u2013 (3y)2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

2(4x + 3y) (4x \u2013 3y)<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(ii) (x \u2013 y)2<\/sup> <\/sub>\u2013 9<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x \u2013 y)2<\/sup> <\/sub>\u2013 9<\/p>\n\n\n\n

(x \u2013 y)2<\/sup> \u2013 32<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x \u2013 y + 3) (x \u2013 y \u2013 3)<\/p>\n\n\n\n

(ii) 9(x + y)2<\/sup> \u2013 x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9[(x + y)2<\/sup> \u2013 x2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

9[(x + y + x) (x + y \u2013 x)]<\/p>\n\n\n\n

So, 9(2x + y) y<\/p>\n\n\n\n

9y(2x + y)<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) 20x2<\/sup> \u2013 45y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

20x2<\/sup> \u2013 45y2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

5(4x2<\/sup> \u2013 9y2<\/sup>)<\/p>\n\n\n\n

5((2x)2<\/sup> \u2013 (3y)2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

5(2x + 3y) (2x \u2013 3y)<\/p>\n\n\n\n

(ii) 9x2<\/sup> \u2013 4(y + 2x)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9x2<\/sup> \u2013 4(y + 2x)2<\/sup><\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

(3x)2<\/sup> \u2013 [2(y + 2x)]2<\/sup><\/p>\n\n\n\n

(3x)2<\/sup> \u2013 (2y + 4x)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(3x + 2y + 4x) (3x \u2013 2y \u2013 4x)<\/p>\n\n\n\n

(7x + 2y) (-x \u2013 2y)<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

(i) 2(x \u2013 2y)2<\/sup> \u2013 50y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2(x \u2013 2y)2<\/sup> \u2013 50y2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2[(x \u2013 2y)2<\/sup> \u2013 25y2<\/sup>]<\/p>\n\n\n\n

2[(x \u2013 2y)2<\/sup> \u2013 (5y)2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

2[(x \u2013 2y + 5y) (x \u2013 2y \u2013 5y)]<\/p>\n\n\n\n

2[(x + 3y) (x \u2013 7y)]<\/p>\n\n\n\n

2(x + 3y) (x \u2013 7y)<\/p>\n\n\n\n

(ii) 32 \u2013 2(x \u2013 4)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

32 \u2013 2(x \u2013 4)2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2[16 \u2013 (x \u2013 4)2<\/sup>]<\/p>\n\n\n\n

2[42<\/sup>\u2013 (x \u2013 4)2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

2[(4 + x \u2013 4) (4 \u2013 x + 4)]<\/p>\n\n\n\n

2[(x) (8 \u2013 x)]<\/p>\n\n\n\n

2x (8 \u2013 x)<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) 108a2<\/sup> \u2013 3(b \u2013 c)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

108a2<\/sup> \u2013 3(b \u2013 c)2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

3[36a2<\/sup> \u2013 (b \u2013 c)2<\/sup>]<\/p>\n\n\n\n

3[(6a)2<\/sup> \u2013 (b \u2013 c)2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

3[(6a + b \u2013 c) (6a \u2013 b + c)]<\/p>\n\n\n\n

(ii) \u03c0a5<\/sup> \u2013 \u03c03<\/sup>ab2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

\u03c0a5<\/sup> \u2013 \u03c03<\/sup>ab2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

\u03c0a(a4<\/sup> \u2013 \u03c02<\/sup>b2<\/sup>)<\/p>\n\n\n\n

\u03c0a((a2<\/sup>)2<\/sup> \u2013 (\u03c0b)2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

\u03c0a(a2<\/sup> + \u03c0b) (a2<\/sup> \u2013 \u03c0b)<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) 50x2<\/sup> \u2013 2(x \u2013 2)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

50x2<\/sup> \u2013 2(x \u2013 2)2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2[25x2<\/sup> \u2013 (x \u2013 2)2<\/sup>]<\/p>\n\n\n\n

2[(5x)2<\/sup> \u2013 (x \u2013 2)2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

2[(5x + x \u2013 2) (5x \u2013 x + 2)]<\/p>\n\n\n\n

2[(6x \u2013 2) (4x + 2)]<\/p>\n\n\n\n

2(6x \u2013 2) (4x + 2)<\/p>\n\n\n\n

(ii) (x \u2013 2)(x + 2) + 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x2<\/sup> \u2013 22<\/sup>) + 3<\/p>\n\n\n\n

X2<\/sup> \u2013 4 + 3<\/p>\n\n\n\n

X2<\/sup> \u2013 1<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

(x + 1) (x \u2013 1)<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) x \u2013 2y \u2013 x2<\/sup> + 4y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x \u2013 2y \u2013 x2<\/sup> + 4y2<\/sup><\/p>\n\n\n\n

x \u2013 2y \u2013 (x2<\/sup> + (2y)2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

x \u2013 2y \u2013 [(x + 2y) (x \u2013 2y)]<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(x \u2013 2y) (1 \u2013 (x + 2y))<\/p>\n\n\n\n

(x \u2013 2y) (1 \u2013 x \u2013 2y)<\/p>\n\n\n\n

(ii) 4a2<\/sup> \u2013 b2<\/sup> + 2a + b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4a2<\/sup> \u2013 b2<\/sup> + 2a + b<\/p>\n\n\n\n

(2a)2<\/sup> \u2013 b2<\/sup> + 2a + b<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

((2a + b) (2a \u2013 b)) + 1(2a + b)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(2a + b) (2a \u2013 b + 1)<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) a(a \u2013 2) \u2013 b(b \u2013 2)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a(a \u2013 2) \u2013 b(b \u2013 2)<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

a2<\/sup> \u2013 2a \u2013 b2<\/sup> \u2013 2b<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

a2<\/sup> \u2013 b2<\/sup> \u2013 2a \u2013 2b<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)[(a + b)(a \u2013 b)] \u2013 2(a \u2013 b)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(a \u2013 b) (a + b \u2013 2)<\/p>\n\n\n\n

(ii) a(a \u2013 1) \u2013 b(b \u2013 1)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a(a \u2013 1) \u2013 b(b \u2013 1)<\/p>\n\n\n\n

Above question can be written as,<\/p>\n\n\n\n

a2<\/sup> \u2013 a \u2013 b2<\/sup> + b<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

a2<\/sup> \u2013 b2<\/sup> \u2013 a + b<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)[(a + b) (a \u2013 b)] \u2013 1 (a \u2013 b)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(a \u2013 b) (a + b \u2013 1)<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

(i) 9 \u2013 x2 <\/sup>+ 2xy \u2013 y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9 \u2013 x2 <\/sup>+ 2xy \u2013 y2<\/sup><\/p>\n\n\n\n

9 \u2013 x2<\/sup> + 2xy \u2013 y2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

9 \u2013 x2<\/sup> + xy + xy \u2013 y2<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

9 \u2013 x2<\/sup> + xy + 3x \u2013 3x + 3y \u2013 3y + xy \u2013 y2<\/sup><\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

9 \u2013 3x + 3y + 3x \u2013 x2<\/sup> + xy + xy \u2013 3y \u2013 y2<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

3(3 \u2013 x + y) + x(3 \u2013 x + y) + y (-3 \u2013 y + x)<\/p>\n\n\n\n

3(3 \u2013 x + y) + x(3 \u2013 x + y) \u2013 y(3 \u2013 x + y)<\/p>\n\n\n\n

(3 \u2013 x + y) (3 + x \u2013 y)<\/p>\n\n\n\n

(ii) 9x4<\/sup> \u2013 (x2<\/sup> + 2x + 1)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9x4<\/sup> \u2013 (x2<\/sup> + 2x + 1)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(3x2<\/sup>)2<\/sup> \u2013 (x + 1)2<\/sup> \u2026 [because (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, (3x2<\/sup> + x + 1) (3x2<\/sup> \u2013 x \u2013 1)<\/p>\n\n\n\n

11.<\/strong><\/p>\n\n\n\n

(i) 9x4<\/sup> \u2013 x2<\/sup> \u2013 12x \u2013 36<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9x4<\/sup> \u2013 x2<\/sup> \u2013 12x \u2013 36<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

9x4<\/sup> \u2013 (x2<\/sup> + 12x + 36)<\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup><\/p>\n\n\n\n

(3x2<\/sup>)2<\/sup> \u2013 (x2<\/sup> + (2 \u00d7 6 \u00d7 x) + 62<\/sup>)<\/p>\n\n\n\n

So, (3x2<\/sup>)2<\/sup> \u2013 (x + 6)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(3x2<\/sup> + x + 6) (3x2<\/sup> \u2013 x \u2013 6)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 5x2<\/sup> \u2013 x + 5<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 5x2<\/sup> \u2013 x + 5<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x2<\/sup>(x \u2013 5) \u2013 1(x \u2013 5)<\/p>\n\n\n\n

(x \u2013 5) (x2<\/sup> \u2013 1)<\/p>\n\n\n\n

(x \u2013 5) (x2<\/sup> \u2013 12<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x \u2013 5) (x + 1) (x \u2013 1)<\/p>\n\n\n\n

12.<\/strong><\/p>\n\n\n\n

(i) a4<\/sup> \u2013 b4<\/sup> + 2b2<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a4<\/sup> \u2013 b4<\/sup> + 2b2<\/sup> \u2013 1<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a4<\/sup> \u2013 (b4<\/sup> \u2013 2b2<\/sup> + 1)<\/p>\n\n\n\n

We know that, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup><\/p>\n\n\n\n

a4<\/sup> \u2013 ((b2<\/sup>)2<\/sup>) \u2013 (2 \u00d7 b2<\/sup> \u00d7 1) + 12<\/sup>)<\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> \u2013 (b2<\/sup> \u2013 1)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup> \u2013 1) (a2<\/sup> \u2013 b2<\/sup> + 1)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 25x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 25x<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x(x2<\/sup> \u2013 25)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x(x2<\/sup> \u2013 52<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

x(x + 5) (x \u2013 5)<\/p>\n\n\n\n

13.<\/strong><\/p>\n\n\n\n

(i) 2x4<\/sup> \u2013 32<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2x4<\/sup> \u2013 32<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

2(x4<\/sup> \u2013 16)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

2((x2<\/sup>)2<\/sup> \u2013 42<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

2(x2<\/sup> + 4) (x2<\/sup> \u2013 4)<\/p>\n\n\n\n

2(x2<\/sup> + 4) (x2<\/sup> \u2013 22<\/sup>)<\/p>\n\n\n\n

2(x2<\/sup> + 4) (x + 2) (x \u2013 2)<\/p>\n\n\n\n

(ii) a2<\/sup>(b + c) \u2013 (b + c)3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a2<\/sup>(b + c) \u2013 (b + c)3<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(b + c) (a2<\/sup> \u2013 (b + c)2<\/sup>)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(b + c) (a + b + c) (a \u2013 b \u2013 c)<\/p>\n\n\n\n

14.<\/strong><\/p>\n\n\n\n

(i) (a + b)3<\/sup> \u2013 a \u2013 b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(a + b)3<\/sup> \u2013 a \u2013 b<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(a + b)3<\/sup> \u2013 (a + b)<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

(a + b) [(a + b)2<\/sup> \u2013 1]<\/p>\n\n\n\n

(a + b) [(a + b)2<\/sup> \u2013 12<\/sup>]<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(a + b) (a + b + 1) (a + b \u2013 1)<\/p>\n\n\n\n

(ii) x2<\/sup> \u2013 2xy + y2<\/sup> \u2013 a2<\/sup> \u2013 2ab \u2013 b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 2xy + y2<\/sup> \u2013 a2<\/sup> \u2013 2ab \u2013 b2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup> \u2013 2xy + y2<\/sup>) \u2013 (a2<\/sup> + 2ab + b2<\/sup>)<\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> and (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup><\/p>\n\n\n\n

(x2<\/sup> \u2013 (2 \u00d7 x \u00d7 y) + y2<\/sup>) \u2013 (a2<\/sup> + (2 \u00d7 a \u00d7 b) + b2<\/sup>)<\/p>\n\n\n\n

(x \u2013 y)2<\/sup> \u2013 (a + b)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)[(x \u2013 y) + (a + b)] [(x \u2013 y) \u2013 (a + b)]<\/p>\n\n\n\n

(x \u2013 y + a + b) (x \u2013 y \u2013 a \u2013 b)<\/p>\n\n\n\n

15.<\/strong><\/p>\n\n\n\n

(i) (a2<\/sup> \u2013 b2<\/sup>) (c2<\/sup> \u2013 d2<\/sup>) \u2013 4abcd<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(a2<\/sup> \u2013 b2<\/sup>) (c2<\/sup> \u2013 d2<\/sup>) \u2013 4abcd<\/p>\n\n\n\n

a2<\/sup>(c2<\/sup> \u2013 d2<\/sup>) \u2013 b2<\/sup> (c2<\/sup> \u2013 d2<\/sup>) \u2013 4abcd<\/p>\n\n\n\n

a2<\/sup>c2<\/sup> \u2013 a2<\/sup>d2<\/sup> \u2013 b2<\/sup>c2<\/sup> + b2<\/sup>d2<\/sup> \u2013 4abcd<\/p>\n\n\n\n

a2<\/sup>c2<\/sup> + b2<\/sup>d2<\/sup> \u2013 a2<\/sup>d2<\/sup> \u2013 b2<\/sup>c2<\/sup> \u2013 2abcd \u2013 2abcd<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

a2<\/sup>c2<\/sup> + b2<\/sup>d2<\/sup> \u2013 2abcd \u2013 a2<\/sup>d2<\/sup> \u2013 b2<\/sup>c2<\/sup> \u2013 2abcd<\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> and (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup><\/p>\n\n\n\n

(ac \u2013 bd)2<\/sup> \u2013 (ad \u2013 bc)2<\/sup><\/p>\n\n\n\n

(ac \u2013 bd + ad \u2013 bc) (ac \u2013 bd \u2013 ad + bc)<\/p>\n\n\n\n

(ii) 4x2<\/sup> \u2013 y2<\/sup> \u2013 3xy + 2x \u2013 2y<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4x2<\/sup> \u2013 y2<\/sup> \u2013 3xy + 2x \u2013 2y<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x2<\/sup> + 3x2<\/sup> \u2013 y2<\/sup> \u2013 3xy + 2x \u2013 2y<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

(x2<\/sup> \u2013 y2<\/sup>) + (3x2<\/sup> \u2013 3xy) + (2x \u2013 2y)<\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b) and take out common terms,<\/p>\n\n\n\n

(x + y) (x \u2013 y) + 3x(x \u2013 y) + 2(x \u2013 y)<\/p>\n\n\n\n

(x \u2013 y) [(x + y) + 3x + 2]<\/p>\n\n\n\n

(x \u2013 y) (x + y + 3x + 2)<\/p>\n\n\n\n

(x \u2013 y) (4x + y + 2)<\/p>\n\n\n\n

16.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + 1\/x2<\/sup> \u2013 11<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> \u2013 11<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x2<\/sup> + (1\/x2<\/sup>) \u2013 2 \u2013 9<\/p>\n\n\n\n

Then, (x2<\/sup> + (1\/x2<\/sup>) \u2013 2) \u2013 32<\/sup><\/p>\n\n\n\n

We know that, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>,<\/p>\n\n\n\n

(x2<\/sup> \u2013 (2 \u00d7 x2<\/sup> \u00d7 (1\/x2<\/sup>)) + (1\/x)2<\/sup>)<\/p>\n\n\n\n

(x \u2013 1\/x)2<\/sup> \u2013 32<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x \u2013 1\/x + 3) (x \u2013 1\/x \u2013 3)<\/p>\n\n\n\n

(ii) x4<\/sup> + 5x2<\/sup> + 9<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x4<\/sup> + 5x2<\/sup> + 9<\/p>\n\n\n\n

x4<\/sup> + 6x2<\/sup> \u2013 x2<\/sup> + 9<\/p>\n\n\n\n

(x4<\/sup> + 6x2<\/sup> + 9) \u2013 x2<\/sup><\/p>\n\n\n\n

((x2<\/sup>)2<\/sup> + (2 \u00d7 x2<\/sup> \u00d7 3) + 32<\/sup>)<\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>,<\/p>\n\n\n\n

((x2<\/sup>)2<\/sup> + (2 \u00d7 x2<\/sup> \u00d7 3) + 32<\/sup>)<\/p>\n\n\n\n

So, (x2<\/sup> + 3)2<\/sup> \u2013 x2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x2<\/sup> + 3 + x) (x2<\/sup> + 3 \u2013 x)<\/p>\n\n\n\n

17.<\/strong><\/p>\n\n\n\n

(i) a4<\/sup> + b4<\/sup> \u2013 7a2<\/sup>b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a4<\/sup> + b4<\/sup> \u2013 7a2<\/sup>b2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a4<\/sup> + b4<\/sup> + 2a2<\/sup>b2<\/sup> \u2013 9a2<\/sup>b2<\/sup><\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>,[(a2<\/sup>)2<\/sup> + (b2<\/sup>)2<\/sup> + (2 \u00d7 a2<\/sup> \u00d7 b2<\/sup>)] \u2013 (3ab)2<\/sup><\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup>)2<\/sup> \u2013 (3ab)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup> + 3ab) (a2<\/sup> + b2<\/sup> \u2013 3ab)<\/p>\n\n\n\n

(ii) x4<\/sup> \u2013 14x2<\/sup> + 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x4<\/sup> \u2013 14x2<\/sup> + 1<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x4<\/sup> + 2x2<\/sup> + 1 \u2013 16x2<\/sup><\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>,<\/p>\n\n\n\n

So, [(x2<\/sup>)2<\/sup> + (2 \u00d7 x2<\/sup> \u00d7 1) + 12<\/sup>] \u2013 16x2<\/sup><\/p>\n\n\n\n

(x2<\/sup> + 1)2<\/sup> \u2013 (4x)2<\/sup><\/p>\n\n\n\n

We know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x2<\/sup> + 1 + 4x) (x2<\/sup> + 1 \u2013 4x)<\/p>\n\n\n\n

18. Express each of the following as the difference of two squares:<\/strong><\/p>\n\n\n\n

(i) (x2<\/sup> \u2013 5x + 7) (x2<\/sup> + 5x + 7)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 5x + 7) (x2<\/sup> + 5x + 7)<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

((x2<\/sup> + 7) \u2013 5x) ((x2<\/sup> + 7) + 5x)<\/p>\n\n\n\n

As, we know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, (x2<\/sup> + 7)2<\/sup> \u2013 (5x)2<\/sup><\/p>\n\n\n\n

(x2<\/sup> + 7)2<\/sup> -25x2<\/sup><\/p>\n\n\n\n

(ii) (x2<\/sup> \u2013 5x + 7) (x2<\/sup> \u2013 5x \u2013 7)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 5x + 7) (x2<\/sup> \u2013 5x \u2013 7)[(x2<\/sup> \u2013 5x) + 7) ((x2<\/sup> \u2013 5x) \u2013 7)<\/p>\n\n\n\n

As, we know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x2<\/sup> \u2013 5x)2<\/sup> \u2013 72<\/sup><\/p>\n\n\n\n

(x2<\/sup> \u2013 5x)2<\/sup> \u2013 49<\/p>\n\n\n\n

(iii) (x2<\/sup> + 5x \u2013 7) (x2<\/sup> \u2013 5x + 7)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> + 5x \u2013 7) (x2<\/sup> \u2013 5x + 7)[x2<\/sup> + (5x \u2013 7)] [x2<\/sup> \u2013 (5x \u2013 7)]<\/p>\n\n\n\n

As, we know that, a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b)<\/p>\n\n\n\n

x2<\/sup> \u2013 (5x \u2013 7)2<\/sup><\/p>\n\n\n\n

We know that, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>,<\/p>\n\n\n\n

X2<\/sup> \u2013 [(5x)2<\/sup> \u2013 (2 \u00d7 5x \u00d7 7) + 72<\/sup>]<\/p>\n\n\n\n

X2<\/sup> \u2013 (25x2<\/sup> \u2013 70x + 49)<\/p>\n\n\n\n

X2<\/sup> \u2013 25x2<\/sup> + 70x \u2013 49<\/p>\n\n\n\n

-24x2<\/sup> + 70x \u2013 49<\/p>\n\n\n\n

19. Evaluate the following by using factors:<\/strong><\/p>\n\n\n\n

(i) (979)2<\/sup> \u2013 (21)2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (99.9)2<\/sup> \u2013 (0.1)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (979)2<\/sup> \u2013 (21)2<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= (979 + 21) (979 \u2013 21)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 1000 \u00d7958<\/p>\n\n\n\n

= 958000<\/p>\n\n\n\n

(ii) (99.9)2<\/sup> \u2013 (0.1)2<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

= (99.9 + 0.1) (99.9 \u2013 0.1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 100 \u00d7 99.8<\/p>\n\n\n\n

= 9980<\/p>\n\n\n\n

Exercise 4.4<\/h4>\n\n\n\n

Factorise the following (1 to 18):<\/strong><\/p>\n\n\n\n

1.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + 5x + 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

\"MLx2<\/sup> + 5x + 6<\/p>\n\n\n\n

x2<\/sup> + 3x + 2x + 6<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x(x + 3) + 2 (x + 3)<\/p>\n\n\n\n

(x + 3) (x + 2)<\/p>\n\n\n\n

(ii) x2<\/sup> \u2013 8x + 7<\/strong><\/p>\n\n\n\n

\"MLSolution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 8x + 7<\/p>\n\n\n\n

x2<\/sup> \u2013 7x \u2013 x + 7<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x(x \u2013 7) \u2013 1(x \u2013 7)<\/p>\n\n\n\n

(x \u2013 7) (x \u2013 1)<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + 6x \u2013 7<\/strong><\/p>\n\n\n\n

\"MLSolution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> + 6x \u2013 7<\/p>\n\n\n\n

x2<\/sup> + 7x \u2013 x \u2013 7<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x(x + 7) \u2013 1(x + 7)<\/p>\n\n\n\n

(x + 7) (x \u2013 1)<\/p>\n\n\n\n

(ii) y2<\/sup> + 7y \u2013 18<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

\"MLy2<\/sup> + 7y \u2013 18<\/p>\n\n\n\n

y2<\/sup> + 9y \u2013 2y \u2013 18<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

y(y + 9) \u2013 2(y + 9)<\/p>\n\n\n\n

(y + 9) (y \u2013 2)<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(i) y2<\/sup> \u2013 7y \u2013 18<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

y2<\/sup> \u2013 7y \u2013 18<\/p>\n\n\n\n

y2<\/sup> + 2y \u2013 9y \u2013 18<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

y(y + 2) \u2013 9(y + 2)<\/p>\n\n\n\n

(y + 2) (y \u2013 9)<\/p>\n\n\n\n

(ii) a2<\/sup> \u2013 3a \u2013 54<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a2<\/sup> \u2013 3a \u2013 54<\/p>\n\n\n\n

a2<\/sup> + 6a \u2013 9a \u2013 54<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

a(a + 6) \u2013 9(a + 6)<\/p>\n\n\n\n

So, (a + 6) (a \u2013 9)<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) 2x2<\/sup> \u2013 7x + 6<\/strong><\/p>\n\n\n\n

\"MLSolution:-<\/strong><\/p>\n\n\n\n

2x2<\/sup> \u2013 7x + 6<\/p>\n\n\n\n

2x2<\/sup> \u2013 4x \u2013 3x + 6<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2x(x \u2013 2) \u2013 3(x \u2013 2)<\/p>\n\n\n\n

(x \u2013 2) (2x \u2013 3)<\/p>\n\n\n\n

(ii) 6x2<\/sup> + 13x \u2013 5<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

\"ML6x2<\/sup> + 13x \u2013 5<\/p>\n\n\n\n

6x2<\/sup> + 15x \u2013 2x \u2013 5<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3x(2x + 5) \u2013 1(2x + 5)<\/p>\n\n\n\n

(2x + 5) (3x \u2013 1)<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

\"ML(i) 6x2<\/sup> + 11x \u2013 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

6x2<\/sup> + 11x \u2013 10<\/p>\n\n\n\n

6x2<\/sup> + 15x \u2013 4x \u2013 10<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3x(2x + 5) \u2013 2(2x + 5)<\/p>\n\n\n\n

(2x + 5) (3x \u2013 2)<\/p>\n\n\n\n

(ii) 6x2<\/sup> \u2013 7x \u2013 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

\"ML6x2<\/sup> \u2013 7x \u2013 3<\/p>\n\n\n\n

6x2<\/sup> \u2013 9x + 2x \u2013 3<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3x(2x \u2013 3) + 1(2x \u2013 3)<\/p>\n\n\n\n

(2x \u2013 3) (3x + 1)<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) 2x2<\/sup> \u2013 x \u2013 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2x2<\/sup> \u2013 x \u2013 6<\/p>\n\n\n\n

2x2<\/sup> \u2013 4x + 3x \u2013 6<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2x(x \u2013 2) + 3(x \u2013 2)<\/p>\n\n\n\n

(x \u2013 2) (2x + 3)<\/p>\n\n\n\n

(ii) 1 \u2013 18y \u2013 63y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

1 \u2013 18y \u2013 63y2<\/sup><\/p>\n\n\n\n

1 \u2013 21y + 3y \u2013 63y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

1(1 \u2013 21y) + 3y(1 \u2013 21y)<\/p>\n\n\n\n

(1 \u2013 21y) (1 + 3y)<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) 2y2<\/sup> + y \u2013 45<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2y2<\/sup> + y \u2013 45<\/p>\n\n\n\n

2y2<\/sup> + 10y \u2013 9y \u2013 45<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2y (y + 5) \u2013 9(y + 5)<\/p>\n\n\n\n

(y + 5) (2y \u2013 9)<\/p>\n\n\n\n

(ii) 5 \u2013 4x \u2013 12x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5 \u2013 4x \u2013 12x2<\/sup><\/p>\n\n\n\n

5 \u2013 10x + 6x \u2013 12x2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

5(1 \u2013 2x) + 6x(1 \u2013 2x)<\/p>\n\n\n\n

(1 \u2013 2x) (5 + 6x)<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) x(12x + 7) \u2013 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x(12x + 7) \u2013 10<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

12x2<\/sup> + 7x \u2013 10<\/p>\n\n\n\n

12x2<\/sup> + 15x \u2013 8x \u2013 10<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3x(4x + 5) \u2013 2(4x + 5)<\/p>\n\n\n\n

(4x + 5) (3x \u2013 2)<\/p>\n\n\n\n

(ii) (4 \u2013 x)2<\/sup> \u2013 2x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(4 \u2013 x)2<\/sup> \u2013 2x<\/p>\n\n\n\n

We know that, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup><\/p>\n\n\n\n

So, (42<\/sup> \u2013 (2 \u00d7 4 \u00d7 x) + x2<\/sup>) \u2013 2x<\/p>\n\n\n\n

16 \u2013 8x + x2<\/sup> \u2013 2x<\/p>\n\n\n\n

x2<\/sup> \u2013 10x + 16<\/p>\n\n\n\n

x2<\/sup> \u2013 8x \u2013 2x + 16<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x(x \u2013 8) \u2013 2(x \u2013 8)<\/p>\n\n\n\n

(x \u2013 8) (x -2)<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) 60x2<\/sup> \u2013 70x \u2013 30<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

60x2<\/sup> \u2013 70x \u2013 30<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

10(6x2<\/sup> \u2013 7x \u2013 3)<\/p>\n\n\n\n

10(6x2<\/sup> \u2013 9x + 2x \u2013 3)<\/p>\n\n\n\n

Again, take out common in all terms we get,<\/p>\n\n\n\n

10(3x(2x \u2013 3) + 1(2x \u2013 3))<\/p>\n\n\n\n

10(2x \u2013 3) (3x + 1)<\/p>\n\n\n\n

(ii) x2<\/sup> \u2013 6xy \u2013 7y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 6xy \u2013 7y2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2013 7xy + xy \u2013 7y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x(x \u2013 7y) + y(x \u2013 7y)<\/p>\n\n\n\n

(x \u2013 7y) (x + y)<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

(i) 2x2<\/sup> + 13xy \u2013 24y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2x2<\/sup> + 13xy \u2013 24y2<\/sup><\/p>\n\n\n\n

2x2<\/sup> + 16xy \u2013 3xy \u2013 24y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2x(x + 8y) \u2013 3y(x + 8y)<\/p>\n\n\n\n

(x + 8y) (2x \u2013 3y)<\/p>\n\n\n\n

(ii) 6x2<\/sup> \u2013 5xy \u2013 6y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

6x2<\/sup> \u2013 5xy \u2013 6y2<\/sup><\/p>\n\n\n\n

6x2<\/sup> \u2013 9xy + 4xy \u2013 6y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3x(2x \u2013 3y) + 2y (2x \u2013 3y)<\/p>\n\n\n\n

(2x \u2013 3y) (3x + 2y)<\/p>\n\n\n\n

11.<\/strong><\/p>\n\n\n\n

(i) 5x2<\/sup> + 17xy \u2013 12y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5x2<\/sup> + 17xy \u2013 12y2<\/sup><\/p>\n\n\n\n

5x2<\/sup> + 20xy \u2013 3xy \u2013 12y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

5x(x + 4y) \u2013 3y(x + 4y)<\/p>\n\n\n\n

(x + 4y) (5x \u2013 3y)<\/p>\n\n\n\n

(ii) x2<\/sup>y2<\/sup> \u2013 8xy \u2013 48<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup>y2<\/sup> \u2013 8xy \u2013 48<\/p>\n\n\n\n

x2<\/sup>y2<\/sup> \u2013 12xy + 4xy \u2013 48<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

xy(xy \u2013 12) + 4(xy \u2013 12)<\/p>\n\n\n\n

(xy \u2013 12) (xy + 4)<\/p>\n\n\n\n

12.<\/strong><\/p>\n\n\n\n

(i) 2a2<\/sup>b2<\/sup> \u2013 7ab \u2013 30<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2a2<\/sup>b2<\/sup> \u2013 7ab \u2013 30<\/p>\n\n\n\n

2a2<\/sup>b2<\/sup> \u2013 12ab + 5ab \u2013 30<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2ab(ab \u2013 6) + 5 (ab \u2013 6)<\/p>\n\n\n\n

(ab \u2013 6) (2ab + 5)<\/p>\n\n\n\n

(ii) a(2a \u2013 b) \u2013 b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a(2a \u2013 b) \u2013 b2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

2a2<\/sup> \u2013 ab \u2013 b2<\/sup><\/p>\n\n\n\n

2a2<\/sup> \u2013 2ab + ab \u2013 b2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2a(a \u2013 b) + b(a \u2013 b)<\/p>\n\n\n\n

(a \u2013 b) (2a + b)<\/p>\n\n\n\n

13.<\/strong><\/p>\n\n\n\n

(i) (x \u2013 y)2<\/sup> \u2013 6(x \u2013 y) + 5<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x \u2013 y)2<\/sup> \u2013 6(x \u2013 y) + 5<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x \u2013 y)2<\/sup> \u2013 5(x \u2013 y) \u2013 (x \u2013 y) + 5<\/p>\n\n\n\n

(x \u2013 y) (x \u2013 y \u2013 5) \u2013 1(x \u2013 y \u2013 5)<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

(x \u2013 y \u2013 5) (x \u2013 y \u2013 1)<\/p>\n\n\n\n

(ii) (2x \u2013 y)2<\/sup> \u2013 11(2x \u2013 y) + 28<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(2x \u2013 y)2<\/sup> \u2013 11(2x \u2013 y) + 28<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(2x \u2013 y)2<\/sup> \u2013 7(2x \u2013 y) \u2013 4(2x \u2013 y) + 28<\/p>\n\n\n\n

(2x \u2013 y) (2x \u2013 y \u2013 7) \u2013 4(2x \u2013 y \u2013 7)<\/p>\n\n\n\n

(2x \u2013 y \u2013 7) (2x \u2013 y \u2013 4)<\/p>\n\n\n\n

14.<\/strong><\/p>\n\n\n\n

(i) 4(a \u2013 1)2<\/sup> \u2013 4(a \u2013 1) \u2013 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4(a \u2013 1)2<\/sup> \u2013 4(a \u2013 1) \u2013 3<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

4(a \u2013 1)2<\/sup> \u2013 6(a \u2013 1) + 2(a \u2013 1) \u2013 3<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2(a \u2013 1) [2(a \u2013 1) \u2013 3] + 1[2(a \u2013 1) \u2013 3]<\/p>\n\n\n\n

(2(a \u2013 1) \u2013 3) (2(a \u2013 1) + 1)<\/p>\n\n\n\n

(2a \u2013 2 \u2013 3) (2a \u2013 2 + 1)<\/p>\n\n\n\n

(2a \u2013 5) (2a \u2013 1)<\/p>\n\n\n\n

(ii) 1 \u2013 2a \u2013 2b \u2013 3(a + b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

1 \u2013 2a \u2013 2b \u2013 3(a + b)2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

1 \u2013 2(a + b) \u2013 3(a + b)2<\/sup><\/p>\n\n\n\n

1 \u2013 3(a + b) + (a + b) \u2013 3(a + b)2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

1(1 \u2013 3(a + b)) + (a + b) (1 \u2013 (a + b))<\/p>\n\n\n\n

(1 \u2013 3(a + b)) (1 + (a + b))<\/p>\n\n\n\n

(1 \u2013 3a + 3b) (1 + a + b)<\/p>\n\n\n\n

15.<\/strong><\/p>\n\n\n\n

(i) 3 \u2013 5a \u2013 5b \u2013 12(a + b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

3 \u2013 5a \u2013 5b \u2013 12(a + b)2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

3 \u2013 5(a + b) \u2013 12(a + b)2<\/sup><\/p>\n\n\n\n

3 \u2013 9(a + b) + 4(a + b) \u2013 12(a + b)2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

3(1 \u2013 3(a + b)) + 4(a + b) (1 \u2013 3(a + b))<\/p>\n\n\n\n

(1 \u2013 3(a + b)) (3 + 4(a + b))<\/p>\n\n\n\n

(1 \u2013 3a \u2013 3b) (3 + 4a + 4b)<\/p>\n\n\n\n

(ii) a4<\/sup> \u2013 11a2<\/sup> + 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a4<\/sup> \u2013 11a2<\/sup> + 10<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a4<\/sup> \u2013 10a2<\/sup> \u2013 a2<\/sup> + 10<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

a2<\/sup> (a2<\/sup> \u2013 10) \u2013 1(a2<\/sup> \u2013 10)<\/p>\n\n\n\n

(a2<\/sup> \u2013 10) (a2<\/sup> \u2013 1)<\/p>\n\n\n\n

16.<\/strong><\/p>\n\n\n\n

(i) (x + 4)2<\/sup> \u2013 5xy -20y \u2013 6y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x + 4)2<\/sup> \u2013 5xy -20y \u2013 6y2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x + 4)2<\/sup> \u2013 5y(x + 4) \u2013 6y2<\/sup><\/p>\n\n\n\n

(x + 4)2<\/sup> \u2013 6y(x + 4) + y(x + 4) \u2013 6y2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

(x + 4) (x + 4 \u2013 6y) + y(x + 4 \u2013 6y)<\/p>\n\n\n\n

(x \u2013 6y + 4) (x + 4 + y)<\/p>\n\n\n\n

(ii) (x2<\/sup> \u2013 2x2<\/sup>) \u2013 23(x2<\/sup> \u2013 2x) + 120<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 2x2<\/sup>) \u2013 23(x2<\/sup> \u2013 2x) + 120<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x)2<\/sup> \u2013 15(x2<\/sup> \u2013 2x) \u2013 8(x2<\/sup> \u2013 2x) + 120<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x) (x2<\/sup> \u2013 2x \u2013 15) \u2013 8(x2<\/sup> \u2013 2x \u2013 15)<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x \u2013 15) (x2<\/sup> \u2013 2x \u2013 8)<\/p>\n\n\n\n

17. 4(2a \u2013 3)2<\/sup> \u2013 3(2a \u2013 3) (a \u2013 1) \u2013 7 (a \u2013 1)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4(2a \u2013 3)2<\/sup> \u2013 3(2a \u2013 3) (a \u2013 1) \u2013 7 (a \u2013 1)2<\/sup><\/p>\n\n\n\n

Let us assume, 2a \u2013 3 = p and a \u2013 1 = q<\/p>\n\n\n\n

So, 4p2<\/sup> \u2013 3pq \u2013 7q2<\/sup><\/p>\n\n\n\n

Then, 4p2<\/sup> \u2013 7pq + 4pq \u2013 7q2<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

P(4p \u2013 7q) + q(4p \u2013 7q)<\/p>\n\n\n\n

(4p \u2013 7q) (p + q)<\/p>\n\n\n\n

Now, substitute the value of p and q we get,<\/p>\n\n\n\n

(4(2a \u2013 3) \u2013 7(a \u2013 1)) (2a \u2013 3 + a \u2013 1)<\/p>\n\n\n\n

(8a \u2013 12 \u2013 7a + 7) (3a \u2013 4)<\/p>\n\n\n\n

(a \u2013 5) (3a \u2013 4)<\/p>\n\n\n\n

18. (2x2<\/sup> + 5x) (2x2<\/sup> + 5x \u2013 19) + 84<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(2x2<\/sup> + 5x) (2x2<\/sup> + 5x \u2013 19) + 84<\/p>\n\n\n\n

Let us assume, 2x2<\/sup> + 5x = p<\/p>\n\n\n\n

So, (p) (p \u2013 19) + 84<\/p>\n\n\n\n

p2<\/sup> \u2013 19p + 84<\/p>\n\n\n\n

p2<\/sup> \u2013 12p \u2013 7p + 84<\/p>\n\n\n\n

p(p \u2013 12) \u2013 7(p \u2013 12)<\/p>\n\n\n\n

(p \u2013 12) (p \u2013 7)<\/p>\n\n\n\n

Now, substitute the value of p we get,<\/p>\n\n\n\n

(2x2<\/sup> + 5x \u2013 12) (2x2<\/sup> + 5x \u2013 7)<\/p>\n\n\n\n

Exercise 4.5<\/h4>\n\n\n\n

Factorise the following (1 to 13):<\/strong><\/p>\n\n\n\n

1.<\/strong><\/p>\n\n\n\n

(i) 8x3<\/sup> + y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

8x3<\/sup> + y3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(2x)3<\/sup> + y3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 2x, b = y<\/p>\n\n\n\n

Then, (2x)3<\/sup> + y3<\/sup> = (2x + y) ((2x)2<\/sup> \u2013 (2x \u00d7 y) + y2<\/sup>)<\/p>\n\n\n\n

= (2x + y) (4x2<\/sup> \u2013 2xy + y2<\/sup>)<\/p>\n\n\n\n

(ii) 64x3<\/sup> \u2013 125y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

64x3<\/sup> \u2013 125y3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(4x)3<\/sup> \u2013 (5y)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 4x, b = 5y<\/p>\n\n\n\n

Then, (4x)3<\/sup> \u2013 (5y)3<\/sup> = (4x \u2013 5y) ((4x)2<\/sup> + (4x \u00d7 5y) + 5y2<\/sup>)<\/p>\n\n\n\n

= (4x \u2013 5y) (16x2<\/sup> + 2oxy + 25y2<\/sup>)<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) 64x3<\/sup> + 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

64x3<\/sup> + 1<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(4x)3<\/sup> + 13<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 4x, b = 1<\/p>\n\n\n\n

Then, (4x)3<\/sup> + 13<\/sup> = (4x + 1) ((4x)2<\/sup> \u2013 (4x \u00d7 1) + 12<\/sup>)<\/p>\n\n\n\n

= (4x + 1) (16x2<\/sup> \u2013 4x + 1)<\/p>\n\n\n\n

(ii) 7a3<\/sup> + 56b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

7a3<\/sup> + 56b3<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

7(a3<\/sup> + 8b3<\/sup>)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

7(a3<\/sup> + (2b)3<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = a, b = 2b<\/p>\n\n\n\n

Then, 7[(a)3<\/sup> + (2b)3<\/sup>] = 7[(a + 2b) ((a)2<\/sup> \u2013 (a \u00d7 2b) + (2b)2<\/sup>)]<\/p>\n\n\n\n

= 7(a + 2b) (a2<\/sup> \u2013 2ab + 4b2<\/sup>)<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(i) (x6<\/sup>\/343) + (343\/x6<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x6<\/sup>\/343) + (343\/x6<\/sup>)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup>\/7)3<\/sup> + (7\/x2<\/sup>)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = (x2<\/sup>\/7), b = (7\/x2<\/sup>)<\/p>\n\n\n\n

Then, (x2<\/sup>\/7)3<\/sup> + (7\/x2<\/sup>)3<\/sup> = [(x2<\/sup>\/7) + (7\/x2<\/sup>)] [(x2<\/sup>\/7)2<\/sup> \u2013 ((x2<\/sup>\/7) \u00d7 (7\/x2<\/sup>)) + (7\/x2<\/sup>)2<\/sup>]<\/p>\n\n\n\n

= [(x2<\/sup>\/7) + (7\/x2<\/sup>)] [(x4<\/sup>\/49) \u2013 1 + (49\/x4<\/sup>)]<\/p>\n\n\n\n

(ii) 8x3<\/sup> \u2013 1\/27y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

8x3<\/sup> \u2013 1\/27y3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(2x)3<\/sup> \u2013 (1\/3y)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 2x, b = (1\/3y)<\/p>\n\n\n\n

Then, (2x)3<\/sup> \u2013 (1\/3y)3<\/sup> = (2x \u2013 (1\/3y)) ((2x)2<\/sup> + (2x \u00d7 (1\/3y)) + (3y)2<\/sup>)<\/p>\n\n\n\n

= (2x \u2013 (1\/3y)) (4x2<\/sup> + (2x\/3y) + 9y2<\/sup>)<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + x5<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup> + x5<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

x2<\/sup>(1 + x3<\/sup>)<\/p>\n\n\n\n

x2<\/sup>(13<\/sup> + x3<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 1, b = x<\/p>\n\n\n\n

= x2<\/sup> [(1 + x) (12<\/sup> \u2013 (1 \u00d7 x) + x2<\/sup>)]<\/p>\n\n\n\n

= x2<\/sup> (1 + x) (1 \u2013 x + x2<\/sup>)<\/p>\n\n\n\n

(ii) 32x4<\/sup> \u2013 500x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

32x4<\/sup> \u2013 500x<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

4x(8x3<\/sup> \u2013 125)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

4x((2x)3<\/sup> \u2013 53<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 2x, b = 5<\/p>\n\n\n\n

= 4x(2x \u2013 5) ((2x)2<\/sup> + (2x \u00d7 5) + 52<\/sup>)<\/p>\n\n\n\n

= 4x(2x \u2013 5) (4x2<\/sup> + 10x + 25)<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

(i) 27x3<\/sup>y3<\/sup> \u2013 8<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

27x3<\/sup>y3<\/sup> \u2013 8<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(3xy)3<\/sup> \u2013 23<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 3xy, b = 2<\/p>\n\n\n\n

= (3xy \u2013 2) ((3xy)2<\/sup> + (3xy \u00d7 2) + 22<\/sup>)<\/p>\n\n\n\n

= (3xy \u2013 2) (9x2<\/sup>y2<\/sup> + 6xy + 4)<\/p>\n\n\n\n

(ii) 27(x + y)3<\/sup> + 8(2x \u2013 y)3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

27(x + y)3<\/sup> + 8(2x \u2013 y)3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

33<\/sup>(x + y)3<\/sup> + 23<\/sup>(2x \u2013 y)3<\/sup><\/p>\n\n\n\n

(3(x + y))3<\/sup> + (2(x \u2013 y))3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = 3(x + y), b = 2(x \u2013 y)<\/p>\n\n\n\n

= [3(x + y) + 2(2x \u2013 y)] [(3(x + y))3<\/sup> \u2013 (3(x + y) \u00d7 2(2x \u2013 y)) + (2(2x \u2013 y))2<\/sup>]<\/p>\n\n\n\n

= [3x + 3y + 4x \u2013 2y] [9(x + y)2<\/sup> \u2013 6(x + y)(2x \u2013 y) + 4(2x \u2013 y)2<\/sup>]<\/p>\n\n\n\n

= (7x \u2013 y) [9(x2<\/sup> + y2<\/sup> + 2xy) \u2013 6(2x2<\/sup> \u2013 xy + 2xy \u2013 y2<\/sup>) + 4(4x2<\/sup> + y2<\/sup> \u2013 4xy)]<\/p>\n\n\n\n

= (7x \u2013 y) [9x2<\/sup> + 9y2<\/sup> + 18xy \u2013 12x2<\/sup> \u2013 6xy \u2013 6y2<\/sup> + 16x2<\/sup> + 4y2<\/sup> \u2013 16xy]<\/p>\n\n\n\n

= (7x \u2013 y) [13x2<\/sup> \u2013 4xy + 19y2<\/sup>]<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) a3<\/sup> + b3<\/sup> + a + b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + a + b<\/p>\n\n\n\n

(a3<\/sup> + b3<\/sup>) + (a + b)<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)[(a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)] + (a + b)<\/p>\n\n\n\n

(a + b) (a2<\/sup> \u2013 ab + b2<\/sup> + 1)<\/p>\n\n\n\n

(ii) a3<\/sup> \u2013 b3<\/sup> \u2013 a + b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> \u2013 a + b<\/p>\n\n\n\n

(a3<\/sup> \u2013 b3<\/sup>) \u2013 (a \u2013 b)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)[(a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)] \u2013 (a \u2013 b)<\/p>\n\n\n\n

(a \u2013 b) (a2<\/sup> + ab + b2<\/sup> \u2013 1)<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) x3<\/sup> + x + 2<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> + x + 2<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x3<\/sup> + x + 1 + 1<\/p>\n\n\n\n

Rearranging the above terms, we get<\/p>\n\n\n\n

(x3<\/sup> + 1) (x + 1)<\/p>\n\n\n\n

(x3<\/sup> + 13<\/sup>) (x + 1)<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)[(x + 1) (x2<\/sup> \u2013 x + 1)] + (x + 1)<\/p>\n\n\n\n

(x + 1) (x2<\/sup> \u2013 x + 1 + 1)<\/p>\n\n\n\n

(x + 1) (x2<\/sup> \u2013 x + 2)<\/p>\n\n\n\n

(ii) a3<\/sup> \u2013 a \u2013 120<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> \u2013 a \u2013 120<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a3<\/sup> \u2013 a \u2013 125 + 5<\/p>\n\n\n\n

Rearranging the above terms, we get<\/p>\n\n\n\n

a3<\/sup> \u2013 125 \u2013 a + 5<\/p>\n\n\n\n

(a3<\/sup> \u2013 125) \u2013 (a \u2013 5)<\/p>\n\n\n\n

(a3<\/sup> \u2013 53<\/sup>) \u2013 (a \u2013 5)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)[(a \u2013 5) (a2<\/sup> + 5a + 52<\/sup>)] \u2013 (a \u2013 5)<\/p>\n\n\n\n

(a \u2013 5) (a2<\/sup> + 5a + 25) \u2013 (a \u2013 5)<\/p>\n\n\n\n

(a \u2013 5) (a2<\/sup> + 5a + 25 \u2013 1)<\/p>\n\n\n\n

(a \u2013 5) (a2<\/sup> + 5a + 24)<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) x3<\/sup> + 6x2<\/sup> + 12x + 16<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> + 6x2<\/sup> + 12x + 16<\/p>\n\n\n\n

x3<\/sup> + 6x2<\/sup> + 12x + 8 + 8<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x3<\/sup> + (3 \u00d7 2 \u00d7 x2<\/sup>) + (3 \u00d7 22<\/sup> \u00d7 x) + 23<\/sup>) + 8<\/p>\n\n\n\n

We know that, (a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup><\/p>\n\n\n\n

Now a = x and b = 2<\/p>\n\n\n\n

So, (x + 2)3<\/sup> + 23<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

(x + 2 + 2) ((x + 2)2<\/sup> \u2013 (2 \u00d7 (x + 2)) + 22<\/sup>)<\/p>\n\n\n\n

(x + 4) (x2<\/sup> + 4 + 4x \u2013 2x \u2013 4 + 4)<\/p>\n\n\n\n

(x + 4) (x2<\/sup> + 2x + 4)<\/p>\n\n\n\n

(ii) a3<\/sup> \u2013 3a2<\/sup>b + 3ab2<\/sup> \u2013 2b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> \u2013 3a2<\/sup>b + 3ab2<\/sup> \u2013 2b3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a3<\/sup> \u2013 3a2<\/sup>b + 3ab2<\/sup> \u2013 b3<\/sup> \u2013 b3<\/sup><\/p>\n\n\n\n

We know that, (a \u2013 b)3<\/sup> = a3<\/sup> \u2013 b3<\/sup> \u2013 3a2<\/sup>b + 3ab2<\/sup><\/p>\n\n\n\n

So, (a \u2013 b)3<\/sup> + b3<\/sup><\/p>\n\n\n\n

We also know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = a \u2013 b, b = b<\/p>\n\n\n\n

(a \u2013 b \u2013 b) ((a \u2013 b)2<\/sup> + (a \u2013 b)b + b2<\/sup>)<\/p>\n\n\n\n

(a \u2013 2b) (a2<\/sup> + b2<\/sup> \u2013 2ab + ab \u2013 b2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

(a \u2013 2b) (a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) 2a3<\/sup> + 16b3<\/sup> \u2013 5a \u2013 10b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2a3<\/sup> + 16b3<\/sup> \u2013 5a \u2013 10b<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

2(a3 <\/sup>+ 8b3<\/sup>) \u2013 5(a + 2b)<\/p>\n\n\n\n

2(a3<\/sup> + (2b)3<\/sup>) \u2013 5(a + 2b)<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

2[(a + 2b) (a2<\/sup> \u2013 2ab + 4b2<\/sup>)] \u2013 5(a + 2b)<\/p>\n\n\n\n

(a + 2b) (2a2<\/sup> \u2013 4ab + 8b2<\/sup> \u2013 5)<\/p>\n\n\n\n

(ii) a3<\/sup> \u2013 (1\/a3<\/sup>) \u2013 2a + 2\/a<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a3<\/sup> \u2013 (1\/a3<\/sup>) \u2013 2a + 2\/a<\/p>\n\n\n\n

(a3<\/sup> \u2013 (1\/a)3<\/sup>) \u2013 2a + 2\/a<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)[(a \u2013 1\/a) \u2013 (a2<\/sup> + (a \u00d7 1\/a) + (1\/a)2<\/sup>] \u2013 2(a \u2013 1\/a)<\/p>\n\n\n\n

(a \u2013 1\/a) (a2<\/sup> + 1 + 1\/a2<\/sup>) \u2013 2(a \u2013 1\/a)<\/p>\n\n\n\n

(a \u2013 1\/a) (a2<\/sup> + 1 + 1\/a2<\/sup> \u2013 2)<\/p>\n\n\n\n

(a \u2013 1\/a) (a2<\/sup> + (1\/a2<\/sup>) \u2013 1)<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

(i) a6<\/sup> \u2013 b6<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a6<\/sup> \u2013 b6<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(a2<\/sup>)3<\/sup> \u2013 (b2<\/sup>)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

So, a = a2<\/sup>, b = b2<\/sup><\/p>\n\n\n\n

(a2<\/sup> \u2013 b2<\/sup>) ((a2<\/sup>)2<\/sup>) + a2<\/sup>b2<\/sup> + (b2<\/sup>)2<\/sup>)<\/p>\n\n\n\n

(a2<\/sup> \u2013 b2<\/sup>) (a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup>)<\/p>\n\n\n\n

(ii) x6<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x6<\/sup> \u2013 1<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup>)3<\/sup> \u2013 13<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

So, a = x2<\/sup>, b = 1<\/p>\n\n\n\n

(x2<\/sup> \u2013 1) ((x2<\/sup>)2<\/sup> + (x2<\/sup> \u00d7 1) + 12<\/sup>)<\/p>\n\n\n\n

(x2<\/sup> \u2013 1) (x4<\/sup> + x2<\/sup> + 1)<\/p>\n\n\n\n

11.<\/strong><\/p>\n\n\n\n

(i) 64x6<\/sup> \u2013 729y6<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

64x6<\/sup> \u2013 729y6<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(2x)6<\/sup> \u2013 (3y)6<\/sup>[(2x)2<\/sup>]3<\/sup> \u2013 [(3y)2<\/sup>]3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

So, a = (2x)2<\/sup>, b = (3y)2<\/sup>[(2x)2<\/sup> \u2013 (3y)2<\/sup>] [((2x)2<\/sup>)2<\/sup> + ((2x)2<\/sup>\u00d7 (3y)2<\/sup>) + ((3y)2<\/sup>)2<\/sup>]<\/p>\n\n\n\n

(4x2<\/sup> \u2013 9y2<\/sup>) [16x4<\/sup> + (4x2<\/sup> \u00d7 9y2<\/sup>) + (9y2<\/sup>)2<\/sup>]<\/p>\n\n\n\n

(4x2<\/sup> \u2013 9y2<\/sup>) [16x4<\/sup> + 36x2<\/sup>y2<\/sup> + 81y4<\/sup>] [(2x)2<\/sup> \u2013 (3y)2<\/sup>] [16x4<\/sup> + 36x2<\/sup>y2<\/sup> + 81y4<\/sup>]<\/p>\n\n\n\n

(2x + 3y) (2x \u2013 3y) (16x4<\/sup> + 36x2<\/sup>y2<\/sup> + 81y4<\/sup>)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 (8\/x)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 (8\/x)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(1\/x) (x3<\/sup> \u2013 8)<\/p>\n\n\n\n

(1\/x) [(x)3<\/sup> \u2013 (2)3<\/sup>]<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

So, a = x, b = 2<\/p>\n\n\n\n

(1\/x) (x \u2013 2) (x2<\/sup> + 2x + 4)<\/p>\n\n\n\n

12.<\/strong><\/p>\n\n\n\n

(i) 250 (a \u2013 b)3<\/sup> + 2<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

250 (a \u2013 b)3<\/sup> + 2<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

2(125(a \u2013 b)3<\/sup> + 1)<\/p>\n\n\n\n

2[(5(a \u2013 b))3<\/sup> + 13<\/sup>]<\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

= 2[(5a \u2013 5b + 1) ((5a \u2013 5b)2<\/sup> \u2013 (5a \u2013 5b)1 + 12<\/sup>)]<\/p>\n\n\n\n

= 2(5a \u2013 5b + 1) (25a2<\/sup> + 25b2<\/sup> \u2013 50ab \u2013 5a + 5b + 1)<\/p>\n\n\n\n

(ii) 32a2<\/sup>x3<\/sup> \u2013 8b2<\/sup>x3<\/sup> \u2013 4a2<\/sup>y3<\/sup> + b2<\/sup>y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

32a2<\/sup>x3<\/sup> \u2013 8b2<\/sup>x3<\/sup> \u2013 4a2<\/sup>y3<\/sup> + b2<\/sup>y3<\/sup><\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

8x3<\/sup>(4a2<\/sup> \u2013 b2<\/sup>) \u2013 y3<\/sup>(4a2<\/sup> \u2013 b2<\/sup>)<\/p>\n\n\n\n

(4a2<\/sup> \u2013 b2<\/sup>) (8x3<\/sup> \u2013 y3<\/sup>)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

((2a)2<\/sup> \u2013 b2<\/sup>) ((2x)3 <\/sup>\u2013 y3<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>) and (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(2a + b) (2a \u2013 b) [(2x \u2013 y) ((2x)2<\/sup> + 2xy + y2<\/sup>)]<\/p>\n\n\n\n

(2a + b) (2a \u2013 b) (2x \u2013 y) (4x2<\/sup> + 2xy + y2<\/sup>)<\/p>\n\n\n\n

13.<\/strong><\/p>\n\n\n\n

(i) x9<\/sup> + y9<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x9<\/sup> + y9<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x3<\/sup>)3<\/sup> + (y3<\/sup>)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

Where, a = x3<\/sup>, b = y3<\/sup><\/p>\n\n\n\n

(x3<\/sup> + y3<\/sup>) ((x3<\/sup>)2<\/sup> \u2013 x3<\/sup>y3<\/sup> + (y3<\/sup>)2<\/sup>)<\/p>\n\n\n\n

(x3<\/sup> + y3<\/sup>) (x6<\/sup> \u2013 x3<\/sup>y3<\/sup> + y6<\/sup>)<\/p>\n\n\n\n

Then, (x3<\/sup> + y3<\/sup>) in the form of (a3<\/sup> + b3<\/sup>)<\/p>\n\n\n\n

(x + y)(x2<\/sup> \u2013 xy + y2<\/sup>) (x6<\/sup> \u2013 x3<\/sup>y3<\/sup> + y6<\/sup>)<\/p>\n\n\n\n

(ii) x6<\/sup> \u2013 7x3<\/sup> \u2013 8<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

X6<\/sup> \u2013 7x3<\/sup> \u2013 8<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup>)3<\/sup> \u2013 7x3<\/sup> \u2013 x3<\/sup> + x3<\/sup> \u2013 8<\/p>\n\n\n\n

(x2<\/sup>)3<\/sup> \u2013 8x3<\/sup> + x3<\/sup> \u2013 23<\/sup><\/p>\n\n\n\n

(((x2<\/sup>)3<\/sup>) \u2013 (2x)3<\/sup>) + (x3<\/sup> \u2013 23<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x) ((x2<\/sup>)2<\/sup> + (x2<\/sup> \u00d7 2x) + (2x)2<\/sup>) + (x \u2013 2) (x2<\/sup> + 2x + 22<\/sup>)<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x) (x4<\/sup> + 2x3<\/sup> + 4x2<\/sup>) + (x \u2013 2) (x2<\/sup> + 2x + 4)<\/p>\n\n\n\n

x(x \u2013 2) x2<\/sup>(x2<\/sup> + 2x + 4) + (x \u2013 2) (x2<\/sup> + 2x + 4)<\/p>\n\n\n\n

Take out common in all terms we get,<\/p>\n\n\n\n

(x \u2013 2) (x2<\/sup> + 2x + 4) ((x \u00d7 x2<\/sup>) + 1)<\/p>\n\n\n\n

(x \u2013 2) (x2<\/sup> + 2x + 4) (x3<\/sup> + 1)<\/p>\n\n\n\n

So, above terms are in the form of a3<\/sup> + b3<\/sup><\/p>\n\n\n\n

Therefore, (x \u2013 2) (x2<\/sup> + 2x + 4) (x + 1) (x2<\/sup> \u2013 x + 1)<\/p>\n\n\n\n

Chapter test<\/p>\n\n\n\n

Factorise the following (1 to 12):<\/strong><\/p>\n\n\n\n

1.<\/strong><\/p>\n\n\n\n

(i) 15(2x \u2013 3)3<\/sup> \u2013 10(2x \u2013 3)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

15(2x \u2013 3)3<\/sup> \u2013 10(2x \u2013 3)<\/strong><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

Then, 5(2x \u2013 3) [3(2x \u2013 3)2<\/sup> \u2013 2]<\/p>\n\n\n\n

(ii) a(b \u2013 c) (b + c) \u2013 d(c \u2013 b)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a(b \u2013 c) (b + c) \u2013 d(c \u2013 b)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a(b \u2013 c) (b + c) + d(b \u2013 c)<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

(b \u2013 c) [a(b + c) + d]<\/p>\n\n\n\n

(b \u2013 c) (ab + ac + d)<\/p>\n\n\n\n

2.<\/strong><\/p>\n\n\n\n

(i) 2a2<\/sup>x \u2013 bx + 2a2<\/sup> \u2013 b<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

2a2<\/sup>x \u2013 bx + 2a2<\/sup> \u2013 b<\/p>\n\n\n\n

Rearrange the above terms we get,<\/p>\n\n\n\n

2a2<\/sup>x + 2a \u2013 bx \u2013 b<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

2a2<\/sup>(x + 1) \u2013 b(x + 1)<\/p>\n\n\n\n

(x + 1) (2a2<\/sup> \u2013 b)<\/p>\n\n\n\n

(ii) p2<\/sup> \u2013 (a + 2b)p + 2ab<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

p2<\/sup> \u2013 (a + 2b)p + 2ab<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

p2<\/sup> \u2013 ap \u2013 2bp + 2ab<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

p(p \u2013 a) \u2013 2b(p \u2013 a)<\/p>\n\n\n\n

(p \u2013 a) (p \u2013 2b)<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

(i) (x2<\/sup> \u2013 y2<\/sup>)z + (y2<\/sup> \u2013 z2<\/sup>)x<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 y2<\/sup>)z + (y2<\/sup> \u2013 z2<\/sup>)x<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

zx2<\/sup> \u2013 zy2<\/sup> + xy2<\/sup> \u2013 xz2<\/sup><\/p>\n\n\n\n

Rearrange the above terms we get,<\/p>\n\n\n\n

zx2<\/sup> \u2013 xz2<\/sup> + xy2<\/sup> \u2013 zy2<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

zx(x \u2013 z) + y2<\/sup>(x \u2013 z)<\/p>\n\n\n\n

(x \u2013 z) (zx + y2<\/sup>)<\/p>\n\n\n\n

(ii) 5a4<\/sup> \u2013 5a3<\/sup> + 30a2<\/sup> \u2013 30a<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

5a4<\/sup> \u2013 5a3<\/sup> + 30a2<\/sup> \u2013 30a<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

5a(a3<\/sup> \u2013 a2<\/sup> + 6a \u2013 6)<\/p>\n\n\n\n

5a[a2<\/sup>(a \u2013 1) + 6(a \u2013 1)]<\/p>\n\n\n\n

5a(a \u2013 1) (a2<\/sup> + 6)<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) b(c -d)2<\/sup> + a(d \u2013 c) + 3c \u2013 3d<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

b(c -d)2<\/sup> + a(d \u2013 c) + 3c \u2013 3d<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

b(c \u2013 d)2<\/sup> \u2013 a(c \u2013 d) + 3c \u2013 3d<\/p>\n\n\n\n

b(c \u2013 d)2<\/sup> \u2013 a(c \u2013 d) + 3(c \u2013 d)<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

(c \u2013 d) [b(c \u2013 d) \u2013 a + 3]<\/p>\n\n\n\n

(c \u2013 d) (bc \u2013 bd \u2013 a + 3)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 x2<\/sup> \u2013 xy + x + y \u2013 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 x2<\/sup> \u2013 xy + x + y \u2013 1<\/p>\n\n\n\n

Rearrange the above terms we get,<\/p>\n\n\n\n

x3<\/sup> \u2013 x2<\/sup> \u2013 xy + y + x \u2013 1<\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

x2<\/sup>(x \u2013 1) \u2013 y(x \u2013 1) + 1(x \u2013 1)<\/p>\n\n\n\n

(x \u2013 1) (x2<\/sup> \u2013 y + 1)<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

(i) x(x + z) \u2013 y (y + z)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x(x + z) \u2013 y (y + z)<\/p>\n\n\n\n

x2<\/sup> + xz \u2013 y2<\/sup> \u2013 yz<\/p>\n\n\n\n

Rearrange the above terms we get,<\/p>\n\n\n\n

x2<\/sup> \u2013 y2<\/sup> + xz \u2013 yz<\/p>\n\n\n\n

We know that, (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

So, (x + y) (x \u2013 y) + z(x \u2013 y)<\/p>\n\n\n\n

(x \u2013 y) (x + y + z)<\/p>\n\n\n\n

(ii) a12<\/sup>x4<\/sup> \u2013 a4<\/sup>x12<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a12<\/sup>x4<\/sup> \u2013 a4<\/sup>x12<\/sup><\/p>\n\n\n\n

Take out common in both terms,<\/p>\n\n\n\n

a4<\/sup>x4<\/sup> (a8<\/sup> \u2013 x8<\/sup>)<\/p>\n\n\n\n

a4<\/sup>x4<\/sup>((a4<\/sup>)2<\/sup> \u2013 (x4<\/sup>)2<\/sup>)<\/p>\n\n\n\n

We know that, (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) (a4<\/sup> \u2013 x4<\/sup>)<\/p>\n\n\n\n

a4<\/sup>x4<\/sup> (a4<\/sup> <\/sub>+ x4<\/sup>) ((a2<\/sup>)2<\/sup> \u2013 (x2<\/sup>)2<\/sup>)<\/p>\n\n\n\n

a4<\/sup>x4<\/sup>(a4<\/sup> + x4<\/sup>) (a2<\/sup> + x2<\/sup>) (a2<\/sup> \u2013 x2<\/sup>)<\/p>\n\n\n\n

a4<\/sup>x4<\/sup> (a4<\/sup> + x4<\/sup>) (a2<\/sup> + x2<\/sup>) (a + x) (a \u2013 x)<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

(i) 9x2<\/sup> + 12x + 4 \u2013 16y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9x2<\/sup> + 12x + 4 \u2013 16y2<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(3x)2<\/sup> + (2 \u00d7 3x \u00d7 2) + 22<\/sup> \u2013 16y2<\/sup><\/p>\n\n\n\n

Then, (3x + 2)2<\/sup> + (4y)2<\/sup><\/p>\n\n\n\n

(3x + 2 + 4y) (3x + 2 \u2013 4y)<\/p>\n\n\n\n

(ii) x4<\/sup> + 3x2<\/sup> + 4<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x4<\/sup> + 3x2<\/sup> + 4<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

(x2<\/sup>)2<\/sup> + 3(x2<\/sup>) + 4<\/p>\n\n\n\n

(x2<\/sup>)2<\/sup> + (2)2<\/sup> + 4x2<\/sup> \u2013 x2<\/sup><\/p>\n\n\n\n

(x2<\/sup> + 2)2<\/sup> \u2013 (x2<\/sup>)<\/p>\n\n\n\n

We know that, (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x2<\/sup> + 2 + x) (x2<\/sup> + 2 \u2013 x)<\/p>\n\n\n\n

(x2<\/sup> + x + 2) (x2<\/sup> \u2013 x + 2)<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

(i) 21x2<\/sup> \u2013 59xy + 40y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

21x2<\/sup> \u2013 59xy + 40y2<\/sup><\/p>\n\n\n\n

By multiplying the first and last term we get, 21 \u00d7 40 = 840<\/p>\n\n\n\n

Then, (-35) \u00d7 (-24) = 840<\/p>\n\n\n\n

So, 21x2<\/sup> \u2013 35xy \u2013 24xy + 40y2<\/sup><\/p>\n\n\n\n

7x(3x \u2013 5y) \u2013 8y(3x \u2013 5y)<\/p>\n\n\n\n

(3x \u2013 5y) (7x \u2013 8y)<\/p>\n\n\n\n

(ii) 4x3<\/sup>y \u2013 44x2<\/sup>y + 112xy<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

4x3<\/sup>y \u2013 44x2<\/sup>y + 112xy<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

4xy(x2<\/sup> \u2013 11x + 28)<\/p>\n\n\n\n

Then, 4xy (x2<\/sup> \u2013 7x \u2013 4x + 28)<\/p>\n\n\n\n

4xy [x(x \u2013 7) \u2013 4(x + 7)]<\/p>\n\n\n\n

4xy (x \u2013 7) (x \u2013 4)<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

(i) x2<\/sup>y2<\/sup> \u2013 xy \u2013 72<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x2<\/sup>y2<\/sup> \u2013 xy \u2013 72<\/p>\n\n\n\n

x2<\/sup>y2<\/sup> \u2013 9xy + 8xy \u2013 72<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

xy(xy \u2013 9) + 8(xy \u2013 9)<\/p>\n\n\n\n

(xy \u2013 9) (xy + 8)<\/p>\n\n\n\n

(ii) 9x3<\/sup>y + 41x2<\/sup>y2<\/sup> + 20xy3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

9x3<\/sup>y + 41x2<\/sup>y2<\/sup> + 20xy3<\/sup><\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

xy(9x2<\/sup> + 41xy + y2<\/sup>)<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

xy (9x2<\/sup> + 36xy + 5xy + 20y2<\/sup>)<\/p>\n\n\n\n

xy [9x(x + 4y) + 5y(x + 4y)]<\/p>\n\n\n\n

xy (x + 4y) (9x + 5y)<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

(i) (3a \u2013 2b)2<\/sup> + 3(3a \u2013 2b) \u2013 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(3a \u2013 2b)2<\/sup> + 3(3a \u2013 2b) \u2013 10<\/p>\n\n\n\n

Let us assume, (3a \u2013 2b) = p<\/p>\n\n\n\n

p2<\/sup> + 3p \u2013 10<\/p>\n\n\n\n

p2<\/sup> + 5p \u2013 2p \u2013 10<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

p(p + 5) \u2013 2(p + 5)<\/p>\n\n\n\n

(p + 5) (p \u2013 2)<\/p>\n\n\n\n

Now, substitute the value of p<\/p>\n\n\n\n

(3a \u2013 2b + 5) (3a \u2013 2b \u2013 2)<\/p>\n\n\n\n

(ii) (x2<\/sup> \u2013 3x) (x2<\/sup> \u2013 3x + 7) + 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 3x) (x2<\/sup> \u2013 3x + 7) + 10<\/p>\n\n\n\n

Let us assume, (x2<\/sup> \u2013 3x) = q<\/p>\n\n\n\n

q (q + 7) + 10<\/p>\n\n\n\n

q2<\/sup> + 7q + 10<\/p>\n\n\n\n

q2<\/sup> + 5q + 2q + 10<\/p>\n\n\n\n

q(q + 5) + 2(q + 5)<\/p>\n\n\n\n

(q + 5) (q + 2)<\/p>\n\n\n\n

Now, substitute the value of q<\/p>\n\n\n\n

(x2<\/sup> \u2013 3x + 5) (x2<\/sup> \u2013 3x + 2)<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

(i) (x2<\/sup> \u2013 x) (4x2<\/sup> \u2013 4x \u2013 5) \u2013 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x2<\/sup> \u2013 x) (4x2<\/sup> \u2013 4x \u2013 5) \u2013 6<\/p>\n\n\n\n

(x2<\/sup> \u2013 x) [(4x2<\/sup> \u2013 4x) \u2013 5] \u2013 6<\/p>\n\n\n\n

(x2<\/sup> \u2013 x) [4(x2<\/sup> \u2013 x) \u2013 5] \u2013 6<\/p>\n\n\n\n

Let us assume x2<\/sup> \u2013 x = q<\/p>\n\n\n\n

So, q[4q \u2013 5] \u2013 6<\/p>\n\n\n\n

4q2<\/sup> \u2013 5q \u2013 6<\/p>\n\n\n\n

4q2<\/sup> \u2013 8q + 3q \u2013 6<\/p>\n\n\n\n

4q(q \u2013 2) + 3(q \u2013 2)<\/p>\n\n\n\n

(q \u2013 2) (4q + 3)<\/p>\n\n\n\n

Now, substitute the value of q<\/p>\n\n\n\n

(x2<\/sup> \u2013 x \u2013 2) (4(x2<\/sup> \u2013 x) + 3)<\/p>\n\n\n\n

(x2<\/sup> \u2013 x \u2013 2) (4x2<\/sup> \u2013 4x + 3)<\/p>\n\n\n\n

(x2<\/sup> \u2013 2x + x \u2013 2) (4x2<\/sup> \u2013 4x + 3)[x(x \u2013 2) + 1(x \u2013 2)] (4x2<\/sup> \u2013 4x + 3)<\/p>\n\n\n\n

(x \u2013 2) (x + 1) (4x2<\/sup> \u2013 4x + 3)<\/p>\n\n\n\n

(ii) x4<\/sup> + 9x2<\/sup>y2<\/sup> + 81y4<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x4<\/sup> + 9x2<\/sup>y2<\/sup> + 81y4<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x4<\/sup> + 18x2<\/sup>y2<\/sup> + 81y4<\/sup> \u2013 9x2<\/sup>y2<\/sup><\/p>\n\n\n\n

((x2<\/sup>)2<\/sup> + (2 \u00d7 x2<\/sup> \u00d7 9y2<\/sup>) + (9y2<\/sup>)2<\/sup>) \u2013 9x2<\/sup>y2<\/sup><\/p>\n\n\n\n

We know that, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup><\/p>\n\n\n\n

(x2<\/sup> + 9y2<\/sup>)2<\/sup> \u2013 (3xy)2<\/sup><\/p>\n\n\n\n

(x2<\/sup> + 9y2<\/sup> + 3xy) (x2<\/sup> + 9y2<\/sup> \u2013 3xy)<\/p>\n\n\n\n

11.<\/strong><\/p>\n\n\n\n

(i) (8\/27)x3<\/sup> \u2013 (1\/8)y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(8\/27)x3<\/sup> \u2013 (1\/8)y3<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

((2\/3)x)3<\/sup> \u2013 (\u00bdy)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>)<\/p>\n\n\n\n

((2\/3)x \u2013 \u00bdy) [(2\/3)x + (2\/3)x (1\/2)y + ((1\/2)y)2<\/sup>]<\/p>\n\n\n\n

((2\/3)x \u2013 (1\/2)y) [(4\/9)x2<\/sup> + (xy\/3) + (y2<\/sup>\/4)]<\/p>\n\n\n\n

(ii) x6<\/sup> + 63x3<\/sup> \u2013 64<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x6<\/sup> + 63x3<\/sup> \u2013 64<\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

x6<\/sup> + 64x3<\/sup> \u2013 x3<\/sup> \u2013 64<\/p>\n\n\n\n

Take out common in all terms,<\/p>\n\n\n\n

x3<\/sup> (x3<\/sup> + 64) \u2013 1(x3<\/sup> + 64)<\/p>\n\n\n\n

(x3<\/sup> + 64) (x3<\/sup> \u2013 1)<\/p>\n\n\n\n

(x3<\/sup> + 43<\/sup>) (x3<\/sup> \u2013 13<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>) and a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

So, (x + 4) [x2<\/sup> \u2013 4x + 42<\/sup>] (x \u2013 1) [x2<\/sup> + x + 12<\/sup>]<\/p>\n\n\n\n

(x + 4) (x2<\/sup> \u2013 4x + 16) (x \u2013 1) (x2<\/sup> + x + 1)<\/p>\n\n\n\n

12.<\/strong><\/p>\n\n\n\n

(i) x3<\/sup> + x2<\/sup> \u2013 (1\/x2<\/sup>) + (1\/x3<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

x3<\/sup> + x2<\/sup> \u2013 (1\/x2<\/sup>) + (1\/x3<\/sup>)<\/p>\n\n\n\n

Rearranging the above terms, we get,<\/p>\n\n\n\n

x3<\/sup> + (1\/x3<\/sup>) + x2<\/sup> \u2013 (1\/x2<\/sup>)<\/p>\n\n\n\n

We know that, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b) (a2<\/sup> + ab + b2<\/sup>) and (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(x + 1\/x) (x2<\/sup> \u2013 1 + 1\/x2<\/sup>) + (x + 1\/x) (x \u2013 1\/x)<\/p>\n\n\n\n

(x + 1\/x) [x2<\/sup> \u2013 1 + 1\/x2<\/sup> + x \u2013 1\/x]<\/p>\n\n\n\n

(ii) (x + 1)6<\/sup> \u2013 (x \u2013 1)6<\/sup><\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(x + 1)6<\/sup> \u2013 (x \u2013 1)6<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

((x + 1)3<\/sup>)2<\/sup> \u2013 ((x \u2013 1)3<\/sup>)2<\/sup><\/p>\n\n\n\n

We know that, (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)[(x + 1)3<\/sup> + (x \u2013 1)3<\/sup>] [(x + 1)3<\/sup> \u2013 (x \u2013 1)3<\/sup>] [(x + 1) + (x \u2013 1)][(x + 1)2<\/sup> \u2013 (x \u2013 1) (x + 1) + (x \u2013 1)2<\/sup>] [(x + 1) \u2013 (x \u2013 1)][(x + 1)2<\/sup> + (x \u2013 1) (x + 1) + (x \u2013 1)2<\/sup>]<\/p>\n\n\n\n

(x + 1 + x \u2013 1) [x2<\/sup> + 2x + 1 \u2013 x2<\/sup> + 1 + x2<\/sup> + 1 \u2013 2x(x + 1) \u2013 x + 1] [x2<\/sup> + 2x + 1 + x2<\/sup> \u2013 1 + x2<\/sup> \u2013 2x + 1]<\/p>\n\n\n\n

By simplifying we get,<\/p>\n\n\n\n

2x(x2<\/sup> + 3) 2(3x2<\/sup> + 1)<\/p>\n\n\n\n

4x(x2<\/sup> + 3) (3x2<\/sup> + 1)<\/p>\n\n\n\n

13. Show that (97)3<\/sup> + (14)3<\/sup> is divisible by 111<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

(97)3<\/sup> + (14)3<\/sup><\/p>\n\n\n\n

We know that, a3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> \u2013 ab + b2<\/sup>)<\/p>\n\n\n\n

So, (97 + 14) [(97)2<\/sup> \u2013 (97 \u00d7 14) + (14)2<\/sup>]<\/p>\n\n\n\n

111 [(97)2<\/sup> \u2013 (97 \u00d7 14) + (14)2<\/sup>]<\/p>\n\n\n\n

Therefore, it is clear that the given expression is divisible by 111.<\/p>\n\n\n\n

14. If a + b = 8 and ab = 15, find the value of a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup><\/p>\n\n\n\n

Above terms can be written as,<\/p>\n\n\n\n

a4<\/sup> + 2a2<\/sup>b2<\/sup> + b4<\/sup> \u2013 a2<\/sup>b2<\/sup><\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> + 2a2<\/sup>b2<\/sup> + (b2<\/sup>)2<\/sup> \u2013 (ab)2<\/sup><\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup>)2<\/sup> \u2013 (ab)2<\/sup><\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup> + ab) (a2<\/sup> + b \u2013 ab)<\/p>\n\n\n\n

a + b = 8, ab = 15<\/p>\n\n\n\n

So, (a + b)2<\/sup> = 82<\/sup><\/p>\n\n\n\n

a2<\/sup> + 2ab + b2<\/sup> = 64<\/p>\n\n\n\n

a2<\/sup> + 2(15) + b2<\/sup> = 64<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 30 = 64<\/p>\n\n\n\n

By transposing,<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 64 \u2013 30<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 34<\/p>\n\n\n\n

Then, a4<\/sup> + a2<\/sup>b2<\/sup> + b4<\/sup><\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup> + ab) (a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

= (34 + 15) (34 \u2013 15)<\/p>\n\n\n\n

= 49 \u00d7 19<\/p>\n\n\n\n

= 931<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 9 Maths Chapter 4- Factorization<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 9 Maths Chapter 4- Factorization PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n