{"id":545390,"date":"2021-10-04T09:48:06","date_gmt":"2021-10-04T09:48:06","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545390"},"modified":"2021-10-05T09:02:35","modified_gmt":"2021-10-05T09:02:35","slug":"rd-sharma-solutions-for-class-9-maths-chapter-4-algebraic-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-4-algebraic-identities\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 4\u2013Algebraic Identities"},"content":{"rendered":"\n

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

RD Sharma Solutions for Class 9 Maths Chapter 4\u2013Algebraic Identities<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 4, Class 9 Maths Chapter 4 solutions<\/p>\n\n\n\n

Exercise 4.1 Page No: 4.6<\/h4>\n\n\n\n

Question 1: Evaluate each of the following using identities:<\/strong><\/p>\n\n\n\n

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

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

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

(iv) (a \u2013 0.1) (a + 0.1)<\/strong><\/p>\n\n\n\n

(v) (1.5.x2<\/sup> \u2013 0.3y2<\/sup>) (1.5x2<\/sup> + 0.3y2<\/sup>)<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>(2x \u2013 1\/x)2<\/sup>[Use identity: (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab ]<\/p>\n\n\n\n

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

= 4x2 <\/sup>+ 1\/x2 <\/sup>\u2013 4<\/p>\n\n\n\n

(ii) <\/strong>(2x + y) (2x \u2013 y)[Use identity: (a \u2013 b)(a + b) = a2<\/sup> \u2013 b2<\/sup> ]<\/p>\n\n\n\n

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

= 4x2 <\/sup>\u2013 y2<\/sup><\/strong><\/p>\n\n\n\n

(iii) <\/strong>(a2<\/sup>b \u2013 b2<\/sup>a)2<\/sup>[Use identity: (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab ]<\/p>\n\n\n\n

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

= a4<\/sup>b 2 <\/sup>+ b4<\/sup>a2 <\/sup>\u2013 2 a3<\/sup>b3<\/sup><\/p>\n\n\n\n

(iv) <\/strong>(a \u2013 0.1) (a + 0.1)[Use identity: (a \u2013 b)(a + b) = a2<\/sup> \u2013 b2<\/sup> ]<\/p>\n\n\n\n

(a \u2013 0.1) (a + 0.1) = (a)2<\/sup> \u2013 (0.1)2<\/sup><\/p>\n\n\n\n

= (a)2<\/sup> \u2013 0.01<\/p>\n\n\n\n

(v) <\/strong>(1.5 x2<\/sup> \u2013 0.3y2<\/sup>) (1.5 x2<\/sup> + 0.3y2<\/sup>)[Use identity: (a \u2013 b)(a + b) = a2<\/sup> \u2013 b2<\/sup> ]<\/p>\n\n\n\n

(1.5 x2<\/sup> \u2013 0.3y2<\/sup>) (1.5x2<\/sup> + 0.3y2<\/sup>) = (1.5 x2<\/sup> ) 2 <\/sup>\u2013 (0.3y2<\/sup>)2<\/sup><\/p>\n\n\n\n

= 2.25 x4 <\/sup>\u2013 0.09y4<\/sup><\/p>\n\n\n\n

Question 2: Evaluate each of the following using identities:<\/strong><\/p>\n\n\n\n

(i) (399)2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (0.98)2<\/sup><\/strong><\/p>\n\n\n\n

(iii) 991 x 1009<\/strong><\/p>\n\n\n\n

(iv) 117 x 83<\/strong><\/p>\n\n\n\n

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

(i)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: Simplify each of the following:<\/strong><\/p>\n\n\n\n

(i) 175 x 175 +2 x 175 x 25 + 25 x 25<\/strong><\/p>\n\n\n\n

(ii) 322 x 322 \u2013 2 x 322 x 22 + 22 x 22<\/strong><\/p>\n\n\n\n

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24<\/strong><\/p>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

(i)<\/strong> 175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2<\/sup> + 2 (175) (25) + (25)2<\/sup><\/p>\n\n\n\n

= (175 + 25)2<\/sup>[Because a2<\/sup>+ b2<\/sup>+2ab = (a+b)2<\/sup> ]<\/p>\n\n\n\n

= (200)2<\/sup><\/p>\n\n\n\n

= 40000<\/p>\n\n\n\n

So, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.<\/p>\n\n\n\n

(ii)<\/strong> 322 x 322 \u2013 2 x 322 x 22 + 22 x 22<\/p>\n\n\n\n

= (322)2<\/sup> \u2013 2 x 322 x 22 + (22)2<\/sup><\/p>\n\n\n\n

= (322 \u2013 22)2<\/sup>[Because a2<\/sup>+ b2<\/sup>-2ab = (a-b)2<\/sup>]<\/p>\n\n\n\n

= (300)2<\/sup><\/p>\n\n\n\n

= 90000<\/p>\n\n\n\n

So, 322 x 322 \u2013 2 x 322 x 22 + 22 x 22= 90000.<\/p>\n\n\n\n

(iii)<\/strong> 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24<\/p>\n\n\n\n

= (0.76) 2<\/sup> + 2 x 0.76 x 0.24 + (0.24) 2<\/sup><\/p>\n\n\n\n

= (0.76+0.24) 2<\/sup>[ Because a2<\/sup>+ b2<\/sup>+2ab = (a+b)2<\/sup>]<\/p>\n\n\n\n

= (1.00)2<\/sup><\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

So, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.<\/p>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 4: If x + 1\/x = 11, find the value of x2<\/sup> +1\/x2<\/sup>.<\/strong><\/p>\n\n\n\n

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

\"\"<\/figure>\n\n\n\n

Question 5: If x \u2013 1\/x = -1, find the value of x2<\/sup> +1\/x2<\/sup>.<\/strong><\/p>\n\n\n\n

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

\"\"<\/figure>\n\n\n\n
\n\n\n\n

Exercise 4.2 Page No: 4.11<\/h4>\n\n\n\n

Question 1: Write the following in the expanded form:<\/strong><\/p>\n\n\n\n

(i) (a + 2b + c)2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (2a \u2212 3b \u2212 c)2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (\u22123x+y+z)2<\/sup><\/strong><\/p>\n\n\n\n

(iv) (m+2n\u22125p)2<\/sup><\/strong><\/p>\n\n\n\n

(v) (2+x\u22122y)2<\/sup><\/strong><\/p>\n\n\n\n

(vi) (a2 <\/sup>+b2 <\/sup>+c2<\/sup>) 2<\/sup><\/strong><\/p>\n\n\n\n

(vii) (ab+bc+ca) 2<\/sup><\/strong><\/p>\n\n\n\n

(viii) (x\/y+y\/z+z\/x)2<\/strong><\/p>\n\n\n\n

(ix) (a\/bc + b\/ac + c\/ab) 2<\/sup><\/strong><\/p>\n\n\n\n

(x) (x+2y+4z) 2<\/sup><\/strong><\/p>\n\n\n\n

(xi) (2x\u2212y+z) 2<\/sup><\/strong><\/p>\n\n\n\n

(xii) (\u22122x+3y+2z) 2<\/sup><\/strong><\/p>\n\n\n\n

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

Using identities:<\/p>\n\n\n\n

(x + y + z)2<\/sup> = x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2xz<\/p>\n\n\n\n

(i)<\/strong> (a + 2b + c)2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + (2b) 2<\/sup> + c2<\/sup> + 2a(2b) + 2ac + 2(2b)c<\/p>\n\n\n\n

= a2<\/sup> + 4b2<\/sup> + c2<\/sup> + 4ab + 2ac + 4bc<\/p>\n\n\n\n

(ii)<\/strong> (2a \u2212 3b \u2212 c)2<\/sup><\/p>\n\n\n\n

= [(2a) + (\u22123b) + (\u2212c)]2<\/sup><\/p>\n\n\n\n

= (2a) 2 <\/sup>+ (\u22123b) 2 <\/sup>+ (\u2212c) 2 <\/sup>+ 2(2a)(\u22123b) + 2(\u22123b)(\u2212c) + 2(2a)(\u2212c)<\/p>\n\n\n\n

= 4a2 <\/sup>+ 9b2 <\/sup>+ c2 <\/sup>\u2212 12ab + 6bc \u2212 4ca<\/p>\n\n\n\n

(iii)<\/strong> (\u22123x+y+z)2<\/sup><\/p>\n\n\n\n

= [(\u22123x) 2 <\/sup>+ y2 <\/sup>+ z2 <\/sup>+ 2(\u22123x)y + 2yz + 2(\u22123x)z<\/p>\n\n\n\n

= 9x2 <\/sup>+ y2 <\/sup>+ z2 <\/sup>\u2212 6xy + 2yz \u2212 6xz<\/p>\n\n\n\n

(iv)<\/strong> (m+2n\u22125p)2<\/sup><\/p>\n\n\n\n

= m2 <\/sup>+ (2n) 2 <\/sup>+ (\u22125p) 2 <\/sup>+ 2m \u00d7 2n + (2\u00d72n\u00d7\u22125p) + 2m \u00d7 \u22125p<\/p>\n\n\n\n

= m2 <\/sup>+ 4n2<\/sup> + 25p2 <\/sup>+ 4mn \u2212 20np \u2212 10pm<\/p>\n\n\n\n

(v)<\/strong> (2+x\u22122y)2<\/sup><\/p>\n\n\n\n

= 22 <\/sup>+ x2 <\/sup>+ (\u22122y) 2 <\/sup>+ 2(2)(x) + 2(x)(\u22122y) + 2(2)(\u22122y)<\/p>\n\n\n\n

= 4 + x2 <\/sup>+ 4y2 <\/sup>+ 4 x \u2212 4xy \u2212 8y<\/p>\n\n\n\n

(vi)<\/strong> (a2 <\/sup>+b2 <\/sup>+c2<\/sup>) 2<\/sup><\/p>\n\n\n\n

= (a2<\/sup>) 2 <\/sup>+ (b2<\/sup>) 2 <\/sup>+ (c2 <\/sup>) 2 <\/sup>+ 2a2 <\/sup>b2 <\/sup>+ 2b2<\/sup>c2 <\/sup>+ 2a2<\/sup>c2<\/sup><\/p>\n\n\n\n

= a4 <\/sup>+ b4 <\/sup>+ c4 <\/sup>+ 2a2<\/sup> b2 <\/sup>+ 2b2 <\/sup>c2 <\/sup>+ 2c2 <\/sup>a2<\/sup><\/p>\n\n\n\n

(vii)<\/strong> (ab+bc+ca) 2<\/sup><\/p>\n\n\n\n

= (ab)2<\/sup> + (bc) 2 <\/sup>+ (ca) 2 <\/sup>+ 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)<\/p>\n\n\n\n

= a2 <\/sup>b2 <\/sup>+ b2<\/sup>c2 <\/sup>+ c2 <\/sup>a2 <\/sup>+ 2(ac)b2 <\/sup>+ 2(ab)(c) 2 <\/sup>+ 2(bc)(a) 2<\/sup><\/p>\n\n\n\n

(viii)<\/strong> (x\/y+y\/z+z\/x)2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ix)<\/strong> (a\/bc + b\/ac + c\/ab) 2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(x)<\/strong> (x+2y+4z) 2<\/sup><\/p>\n\n\n\n

= x2 <\/sup>+ (2y) 2 <\/sup>+ (4z) 2 <\/sup>+ (2x)(2y) + 2(2y)(4z) + 2x(4z)<\/p>\n\n\n\n

= x2 <\/sup>+ 4y2 <\/sup>+ 16z2 <\/sup>+ 4xy + 16yz + 8xz<\/p>\n\n\n\n

(xi)<\/strong> (2x\u2212y+z) 2<\/sup><\/p>\n\n\n\n

= (2x) 2 <\/sup>+ (\u2212y) 2 <\/sup>+ (z) 2 <\/sup>+ 2(2x)(\u2212y) + 2(\u2212y)(z) + 2(2x)(z)<\/p>\n\n\n\n

= 4x2 <\/sup>+ y2 <\/sup>+ z2 <\/sup>\u2212 4xy\u22122yz+4xz<\/p>\n\n\n\n

(xii)<\/strong> (\u22122x+3y+2z) 2<\/sup><\/p>\n\n\n\n

= (\u22122x) 2 <\/sup>+ (3y) 2 <\/sup>+ ( 2z) 2 <\/sup>+ 2(\u22122x)(3y)+2(3y)(2z)+2(\u22122x)(2z)<\/p>\n\n\n\n

= 4x2 <\/sup>+ 9y2 <\/sup>+ 4z2 <\/sup>\u221212xy+12yz\u22128xz<\/p>\n\n\n\n

Question 2: Simplify<\/strong><\/p>\n\n\n\n

(i) (a + b + c)2<\/sup> + (a \u2212 b + c) 2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (a + b + c)2<\/sup> \u2212 (a \u2212 b + c) 2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (a + b + c)2<\/sup> + (a \u2013 b + c) 2<\/sup> + (a + b \u2212 c) 2<\/sup><\/strong><\/p>\n\n\n\n

(iv) (2x + p \u2212 c)2<\/sup> \u2212 (2x \u2212 p + c) 2<\/sup><\/strong><\/p>\n\n\n\n

(v) (x2<\/sup> + y2<\/sup> \u2212 z2<\/sup>) 2<\/sup> \u2212 (x2<\/sup> \u2212 y2<\/sup> + z2<\/sup>) 2<\/sup><\/strong><\/p>\n\n\n\n

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

(i)<\/strong> (a + b + c)2<\/sup> + (a \u2212 b + c) 2<\/sup><\/p>\n\n\n\n

= (a2 <\/sup>+ b2 <\/sup>+ c2 <\/sup>+ 2ab+2bc+2ca) + (a2 <\/sup>+ (\u2212b) 2 <\/sup>+ c2 <\/sup>\u22122ab\u22122bc+2ca)<\/p>\n\n\n\n

= 2a2 <\/sup>+ 2 b2 <\/sup>+ 2c2 <\/sup>+ 4ca<\/p>\n\n\n\n

(ii)<\/strong> (a + b + c)2<\/sup> \u2212 (a \u2212 b + c) 2<\/sup><\/p>\n\n\n\n

= (a2 <\/sup>+ b2 <\/sup>+ c2 <\/sup>+ 2ab+2bc+2ca) \u2212 (a2 <\/sup>+ (\u2212b) 2 <\/sup>+ c2 <\/sup>\u22122ab\u22122bc+2ca)<\/p>\n\n\n\n

= a2 <\/sup>+ b2 <\/sup>+ c2 <\/sup>+ 2ab + 2bc + 2ca \u2212 a2 <\/sup>\u2212 b2 <\/sup>\u2212 c2 <\/sup>+ 2ab + 2bc \u2212 2ca<\/p>\n\n\n\n

= 4ab + 4bc<\/p>\n\n\n\n

(iii)<\/strong> (a + b + c)2<\/sup> + (a \u2013 b + c) 2<\/sup> + (a + b \u2212 c) 2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca + (a2<\/sup> + b2<\/sup> + (c) 2<\/sup> \u2212 2ab \u2212 2cb + 2ca) + (a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab \u2212 2bc \u2013 2ca)<\/p>\n\n\n\n

= 3 a2<\/sup> + 3b2<\/sup> + 3c2<\/sup> + 2ab \u2212 2bc + 2ca<\/p>\n\n\n\n

(iv)<\/strong> (2x + p \u2212 c)2<\/sup> \u2212 (2x \u2212 p + c) 2<\/sup><\/p>\n\n\n\n

= [4x2<\/sup> + p2<\/sup> + c2<\/sup> + 4xp \u2212 2pc \u2212 4xc] \u2212 [4x2<\/sup> + p2<\/sup> + c2<\/sup> \u2212 4xp\u2212 2pc + 4xc]<\/p>\n\n\n\n

= 4x2<\/sup> + p2<\/sup> + c2<\/sup> + 4xp \u2212 2pc \u2212 4cx \u2212 4x2<\/sup> \u2212 p2<\/sup> \u2212 c2<\/sup> + 4xp + 2pc\u2212 4cx<\/p>\n\n\n\n

= 8xp \u2212 8xc<\/p>\n\n\n\n

= 8(xp \u2212 xc)<\/p>\n\n\n\n

(v)<\/strong> (x2<\/sup> + y2<\/sup> \u2212 z2<\/sup>) 2<\/sup> \u2212 (x2<\/sup> \u2212 y2<\/sup> + z2<\/sup>) 2<\/sup><\/p>\n\n\n\n

= (x2<\/sup> + y2<\/sup> + (\u2212z) 2<\/sup>) 2<\/sup> \u2212 (x2<\/sup> \u2212 y2<\/sup> + z2<\/sup>) 2<\/sup><\/p>\n\n\n\n

= [x4<\/sup> + y4<\/sup> + z4<\/sup> + 2x2<\/sup>y2<\/sup> \u2013 2y2<\/sup>z 2<\/sup> \u2013 2x2<\/sup>z2<\/sup> \u2212 [x4<\/sup> + y4<\/sup> + z4<\/sup> \u2212 2x2<\/sup>y2<\/sup> \u2212 2y2<\/sup>z2<\/sup> + 2x2<\/sup>z2<\/sup>]<\/p>\n\n\n\n

= 4x2<\/sup>y2 <\/sup>\u2013 4z2<\/sup>x2<\/sup><\/p>\n\n\n\n

Question 3: If a + b + c = 0 and a2<\/sup> + b2<\/sup> + c2<\/sup> = 16, find the value of ab + bc + ca.<\/strong><\/p>\n\n\n\n

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

a + b + c = 0 and a2<\/sup> + b2<\/sup> + c2<\/sup> = 16 (given)<\/p>\n\n\n\n

Choose a + b + c = 0<\/p>\n\n\n\n

Squaring both sides,<\/p>\n\n\n\n

(a + b + c)2<\/sup> = 0<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + c2<\/sup> + 2(ab + bc + ca) = 0<\/p>\n\n\n\n

16 + 2(ab + bc + c) = 0<\/p>\n\n\n\n

2(ab + bc + ca) = -16<\/p>\n\n\n\n

ab + bc + ca = -16\/2 = -8<\/p>\n\n\n\n

or ab + bc + ca = -8<\/p>\n\n\n\n

\n\n\n\n

Exercise 4.3 Page No: 4.19<\/h4>\n\n\n\n

Question 1: Find the cube of each of the following binomial expressions:<\/strong><\/p>\n\n\n\n

(i) (1\/x + y\/3)<\/strong><\/p>\n\n\n\n

(ii) (3\/x \u2013 2\/x2<\/sup>)<\/strong><\/p>\n\n\n\n

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

(iv) (4 \u2013 1\/3x)<\/strong><\/p>\n\n\n\n

Solution:<\/strong>[Using identities: (a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b) and (a \u2013 b)3<\/sup> = a3<\/sup> \u2013 b3<\/sup> \u2013 3ab(a \u2013 b) ]<\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 2: Simplify each of the following:<\/strong><\/p>\n\n\n\n

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

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

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

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

Solution:<\/strong>[Using identities:<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab)<\/p>\n\n\n\n

(a + b)(a-b) = a2<\/sup> \u2013 b2<\/sup><\/p>\n\n\n\n

(a + b)2 <\/sup>= a2<\/sup> + b2<\/sup> + 2ab and<\/p>\n\n\n\n

(a \u2013 b)2 <\/sup>= a2<\/sup> + b2<\/sup> \u2013 2ab]<\/p>\n\n\n\n

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

Here a = (x + 3), b = (x \u2013 3)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

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

\"\"<\/figure>\n\n\n\n

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

Here a = (x + 2\/x) and b = (x \u2013 2\/x)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

Here a = (2x \u2013 5y) and b = 2x + 5y<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: If a + b = 10 and ab = 21, find the value of a3<\/sup> + b3<\/sup>.<\/strong><\/p>\n\n\n\n

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

a + b = 10, ab = 21 (given)<\/p>\n\n\n\n

Choose a + b = 10<\/p>\n\n\n\n

Cubing both sides,<\/p>\n\n\n\n

(a + b)3<\/sup> = (10)3<\/sup><\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 3ab(a + b) = 1000<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 3 x 21 x 10 = 1000 (using given values)<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 630 = 1000<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> = 1000 \u2013 630 = 370<\/p>\n\n\n\n

or a3<\/sup> + b3<\/sup> = 370<\/p>\n\n\n\n

Question 4: If a \u2013 b = 4 and ab = 21, find the value of a3<\/sup> \u2013 b3<\/sup>.<\/strong><\/p>\n\n\n\n

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

a \u2013 b = 4, ab= 21 (given)<\/p>\n\n\n\n

Choose a \u2013 b = 4<\/p>\n\n\n\n

Cubing both sides,<\/p>\n\n\n\n

(a \u2013 b)3<\/sup> = (4)3<\/sup><\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> \u2013 3ab (a \u2013 b) = 64<\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> \u2013 3 \u00d7 21 x 4 = 64 (using given values)<\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> \u2013 252 = 64<\/p>\n\n\n\n

a3<\/sup> \u2013 b3<\/sup> = 64 + 252<\/p>\n\n\n\n

= 316<\/p>\n\n\n\n

Or a3<\/sup> \u2013 b3<\/sup> = 316<\/p>\n\n\n\n

Question 5: If x + 1\/x = 5, find the value of x3<\/sup> + 1\/x3 <\/sup>.<\/strong><\/p>\n\n\n\n

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

Given: x + 1\/x = 5<\/p>\n\n\n\n

Apply Cube on x + 1\/x<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 6: If x \u2013 1\/x = 7, find the value of x3<\/sup> \u2013 1\/x3 <\/sup>.<\/strong><\/p>\n\n\n\n

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

Given: x \u2013 1\/x = 7<\/p>\n\n\n\n

Apply Cube on x \u2013 1\/x<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 7: If x \u2013 1\/x = 5, find the value of x3<\/sup> \u2013 1\/x3 <\/sup>.<\/strong><\/p>\n\n\n\n

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

Given: x \u2013 1\/x = 5<\/p>\n\n\n\n

Apply Cube on x \u2013 1\/x<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 8: If (x2<\/sup> + 1\/x2<\/sup>) = 51, find the value of x3<\/sup> \u2013 1\/x3<\/sup>.<\/strong><\/p>\n\n\n\n

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

We know that: (x \u2013 y)2<\/sup> = x2<\/sup> + y2<\/sup> \u2013 2xy<\/p>\n\n\n\n

Replace y with 1\/x, we get<\/p>\n\n\n\n

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

Since (x2<\/sup> + 1\/x2<\/sup>) = 51 (given)<\/p>\n\n\n\n

(x \u2013 1\/x)2<\/sup> = 51 \u2013 2 = 49<\/p>\n\n\n\n

or (x \u2013 1\/x) = \u00b17<\/p>\n\n\n\n

Now, Find x3<\/sup> \u2013 1\/x3<\/sup><\/p>\n\n\n\n

We know that, x3<\/sup> \u2013 y3<\/sup> = (x \u2013 y)(x2<\/sup> + y2<\/sup> + xy)<\/p>\n\n\n\n

Replace y with 1\/x, we get<\/p>\n\n\n\n

x3<\/sup> \u2013 1\/x3<\/sup> = (x \u2013 1\/x)(x2<\/sup> + 1\/x2<\/sup> + 1)<\/p>\n\n\n\n

Use (x \u2013 1\/x) = 7 and (x2<\/sup> + 1\/x2<\/sup>) = 51<\/p>\n\n\n\n

x3<\/sup> \u2013 1\/x3<\/sup> = 7 x 52 = 364<\/p>\n\n\n\n

x3<\/sup> \u2013 1\/x3<\/sup> = 364<\/p>\n\n\n\n

Question 9: If (x2<\/sup> + 1\/x2<\/sup>) = 98, find the value of x3<\/sup> + 1\/x3<\/sup>.<\/strong><\/p>\n\n\n\n

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

We know that: (x + y)2<\/sup> = x2<\/sup> + y2<\/sup> + 2xy<\/p>\n\n\n\n

Replace y with 1\/x, we get<\/p>\n\n\n\n

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

Since (x2<\/sup> + 1\/x2<\/sup>) = 98 (given)<\/p>\n\n\n\n

(x + 1\/x)2<\/sup> = 98 + 2 = 100<\/p>\n\n\n\n

or (x + 1\/x) = \u00b110<\/p>\n\n\n\n

Now, Find x3<\/sup> + 1\/x3<\/sup><\/p>\n\n\n\n

We know that, x3<\/sup> + y3<\/sup> = (x + y)(x2<\/sup> + y2<\/sup> \u2013 xy)<\/p>\n\n\n\n

Replace y with 1\/x, we get<\/p>\n\n\n\n

x3<\/sup> + 1\/x3<\/sup> = (x + 1\/x)(x2<\/sup> + 1\/x2<\/sup> \u2013 1)<\/p>\n\n\n\n

Use (x + 1\/x) = 10 and (x2<\/sup> + 1\/x2<\/sup>) = 98<\/p>\n\n\n\n

x3<\/sup> + 1\/x3<\/sup> = 10 x 97 = 970<\/p>\n\n\n\n

x3<\/sup> + 1\/x3<\/sup> = 970<\/p>\n\n\n\n

Question 10: If 2x + 3y = 13 and xy = 6, find the value of 8x3<\/sup> + 27y3<\/sup>.<\/strong><\/p>\n\n\n\n

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

Given: 2x + 3y = 13, xy = 6<\/p>\n\n\n\n

Cubing 2x + 3y = 13 both sides, we get<\/p>\n\n\n\n

(2x + 3y)3<\/sup> = (13)3<\/sup><\/p>\n\n\n\n

(2x)3<\/sup> + (3y) 3<\/sup> + 3( 2x )(3y) (2x + 3y) = 2197<\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> + 18xy(2x + 3y) = 2197<\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> + 18 x 6 x 13 = 2197<\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> + 1404 = 2197<\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> = 2197 \u2013 1404 = 793<\/p>\n\n\n\n

8x3<\/sup> + 27y3<\/sup> = 793<\/p>\n\n\n\n

Question 11: If 3x \u2013 2y= 11 and xy = 12, find the value of 27x3<\/sup> \u2013 8y3<\/sup>.<\/strong><\/p>\n\n\n\n

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

Given: 3x \u2013 2y = 11 and xy = 12<\/p>\n\n\n\n

Cubing 3x \u2013 2y = 11 both sides, we get<\/p>\n\n\n\n

(3x \u2013 2y)3<\/sup> = (11)3<\/sup><\/p>\n\n\n\n

(3x)3<\/sup> \u2013 (2y)3<\/sup> \u2013 3 ( 3x)( 2y) (3x \u2013 2y) =1331<\/p>\n\n\n\n

27x3<\/sup> \u2013 8y3<\/sup> \u2013 18xy(3x -2y) =1331<\/p>\n\n\n\n

27x3<\/sup> \u2013 8y3<\/sup> \u2013 18 x 12 x 11 = 1331<\/p>\n\n\n\n

27x3<\/sup> \u2013 8y3<\/sup> \u2013 2376 = 1331<\/p>\n\n\n\n

27x3<\/sup> \u2013 8y3<\/sup> = 1331 + 2376 = 3707<\/p>\n\n\n\n

27x3<\/sup> \u2013 8y3<\/sup> = 3707<\/p>\n\n\n\n

\n\n\n\n

Exercise 4.4 Page No: 4.23<\/h4>\n\n\n\n

Question 1: Find the following products:<\/strong><\/p>\n\n\n\n

(i) (3x + 2y)(9x2<\/sup> \u2013 6xy + 4y2<\/sup>)<\/strong><\/p>\n\n\n\n

(ii) (4x \u2013 5y)(16x2<\/sup> + 20xy + 25y2<\/sup>)<\/strong><\/p>\n\n\n\n

(iii) (7p4<\/sup> + q)(49p8<\/sup> \u2013 7p4<\/sup>q + q2<\/sup>)<\/strong><\/p>\n\n\n\n

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

(v) (3\/x \u2013 5\/y)(9\/x2<\/sup> + 25\/y2<\/sup> + 15\/xy)<\/strong><\/p>\n\n\n\n

(vi) (3 + 5\/x)(9 \u2013 15\/x + 25\/x2<\/sup>)<\/strong><\/p>\n\n\n\n

(vii) (2\/x + 3x)(4\/x2 <\/sup>+ 9x2 <\/sup>\u2013 6)<\/strong><\/p>\n\n\n\n

(viii) (3\/x \u2013 2x2<\/sup>)(9\/x2 <\/sup>+ 4x4<\/sup> \u2013 6x)<\/strong><\/p>\n\n\n\n

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

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

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

(xii) (x3 <\/sup>+ 1)(x6<\/sup> \u2013 x3<\/sup> + 1)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> (3x + 2y)(9x2<\/sup> \u2013 6xy + 4y2<\/sup>)<\/p>\n\n\n\n

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

We know, a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

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

= 27x3<\/sup> + 8y3<\/sup><\/p>\n\n\n\n

(ii)<\/strong> (4x \u2013 5y)(16x2<\/sup> + 20xy + 25y2<\/sup>)<\/p>\n\n\n\n

= (4x \u2013 5y)[(4x)2<\/sup> + (4x)(5y) + (5y)2<\/sup>)]<\/p>\n\n\n\n

We know, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab)<\/p>\n\n\n\n

= (4x)3<\/sup> \u2013 (5y) 3<\/sup><\/p>\n\n\n\n

= 64x3<\/sup> \u2013 125y3<\/sup><\/p>\n\n\n\n

(iii)<\/strong> (7p4<\/sup> + q)(49p8<\/sup> \u2013 7p4<\/sup>q + q2<\/sup>)<\/p>\n\n\n\n

= (7p4<\/sup> + q)[(7p4<\/sup>)2<\/sup> \u2013 (7p4<\/sup>)(q) + (q)2<\/sup>)]<\/p>\n\n\n\n

We know, a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)<\/p>\n\n\n\n

= (7p4<\/sup>)3<\/sup> + (q) 3<\/sup><\/p>\n\n\n\n

= 343 p12<\/sup> + q3<\/sup><\/p>\n\n\n\n

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

We know, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab)<\/p>\n\n\n\n

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

\"\"<\/figure>\n\n\n\n

(v)<\/strong> (3\/x \u2013 5\/y)(9\/x2<\/sup> + 25\/y2<\/sup> + 15\/xy)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Using a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab) ]<\/p>\n\n\n\n

(vi)<\/strong> (3 + 5\/x)(9 \u2013 15\/x + 25\/x2<\/sup>)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Using: a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)]<\/p>\n\n\n\n

(vii)<\/strong> (2\/x + 3x)(4\/x2 <\/sup>+ 9x2 <\/sup>\u2013 6)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Using: a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)]<\/p>\n\n\n\n

(viii)<\/strong> (3\/x \u2013 2x2<\/sup>)(9\/x2 <\/sup>+ 4x4<\/sup> \u2013 6x)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Using : a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab)]<\/p>\n\n\n\n

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

And we know, a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab)<\/p>\n\n\n\n

(1 \u2013 x)(1 + x + x2<\/sup>) can be written as<\/p>\n\n\n\n

(1 \u2013 x)[(12<\/sup> + (1)(x)+ x2<\/sup>)]<\/p>\n\n\n\n

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

= 1 \u2013 x3<\/sup><\/p>\n\n\n\n

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

And we know, a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab)]<\/p>\n\n\n\n

(1 + x)(1 \u2013 x + x2<\/sup>) can be written as,<\/p>\n\n\n\n

(1 + x)[(12<\/sup> \u2013 (1)(x) + x2<\/sup>)]<\/p>\n\n\n\n

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

= 1 + x3<\/sup><\/p>\n\n\n\n

(xi)<\/strong> (x2 <\/sup>\u2013 1)(x4<\/sup> + x2 <\/sup>+1) can be written as,<\/p>\n\n\n\n

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

= (x2<\/sup>)3<\/sup> \u2013 13<\/sup><\/p>\n\n\n\n

= x6<\/sup> \u2013 1[using a3<\/sup> \u2013 b3<\/sup> = (a \u2013 b)(a2<\/sup> + b2<\/sup> + ab) ]<\/p>\n\n\n\n

(xii)<\/strong> (x3 <\/sup>+ 1)(x6<\/sup> \u2013 x3<\/sup> + 1) can be written as,<\/p>\n\n\n\n

(x3<\/sup> + 1)[(x3<\/sup>)2<\/sup> \u2013 (x3<\/sup>)(1) + 12<\/sup>]<\/p>\n\n\n\n

= (x3<\/sup>) 3<\/sup> + 13<\/sup><\/p>\n\n\n\n

= x9<\/sup> + 1[using a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> + b2<\/sup> \u2013 ab) ]<\/p>\n\n\n\n

Question 2: If x = 3 and y = -1, find the values of each of the following using in identity:<\/strong><\/p>\n\n\n\n

(i) (9y2<\/sup> \u2013 4x2<\/sup>)(81y4<\/sup> + 36x2<\/sup>y2<\/sup> + 16x4<\/sup>)<\/strong><\/p>\n\n\n\n

(ii) (3\/x \u2013 x\/3)(x2<\/sup> \/9 + 9\/x2<\/sup> + 1)<\/strong><\/p>\n\n\n\n

(iii) (x\/7 + y\/3)(x2<\/sup>\/49 + y2<\/sup>\/9 \u2013 xy\/21)<\/strong><\/p>\n\n\n\n

(iv) (x\/4 \u2013 y\/3)(x2<\/sup>\/16 + xy\/12 + y2<\/sup>\/9)
(v) (5\/x + 5x)(25\/x2<\/sup> \u2013 25 + 25x2<\/sup>)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> (9y2<\/sup> \u2013 4x2<\/sup>)(81y4<\/sup> + 36x2<\/sup>y2<\/sup> + 16x4<\/sup>)<\/p>\n\n\n\n

= (9y2<\/sup> \u2013 4x2<\/sup>) [(9y2<\/sup> ) 2 <\/sup>+ 9y2<\/sup> x 4x2 <\/sup>+ (4x2<\/sup>) 2<\/sup> ]<\/p>\n\n\n\n

= (9y2<\/sup> ) 3 <\/sup>\u2013 (4x2<\/sup>)3<\/sup><\/p>\n\n\n\n

= 729 y6<\/sup> \u2013 64 x6<\/sup><\/p>\n\n\n\n

Put x = 3 and y = -1<\/p>\n\n\n\n

= 729 \u2013 46656<\/p>\n\n\n\n

= \u2013 45927<\/p>\n\n\n\n

(ii)<\/strong> Put x = 3 and y = -1<\/p>\n\n\n\n

(3\/x \u2013 x\/3)(x2<\/sup> \/9 + 9\/x2<\/sup> + 1)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii)<\/strong> Put x = 3 and y = -1<\/p>\n\n\n\n

(x\/7 + y\/3)(x2<\/sup>\/49 + y2<\/sup>\/9 \u2013 xy\/21)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv)<\/strong> Put x = 3 and y = -1<\/p>\n\n\n\n

(x\/4 \u2013 y\/3)(x2<\/sup>\/16 + xy\/12 + y2<\/sup>\/9)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v)<\/strong> Put x = 3 and y = -1<\/p>\n\n\n\n

(5\/x + 5x)(25\/x2<\/sup> \u2013 25 + 25x2<\/sup>)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: If a + b = 10 and ab = 16, find the value of a2<\/sup> \u2013 ab + b2<\/sup> and a2<\/sup> + ab + b2<\/sup>.<\/strong><\/p>\n\n\n\n

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

a + b = 10, ab = 16<\/p>\n\n\n\n

Squaring, a + b = 10, both sides<\/p>\n\n\n\n

(a + b)2<\/sup> = (10)2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2ab = 100<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2 x 16 = 100<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 32 = 100<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 100 \u2013 32 = 68<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 68<\/p>\n\n\n\n

Again, a2<\/sup> \u2013 ab + b2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 ab = 68 \u2013 16 = 52 and<\/p>\n\n\n\n

a2<\/sup> + ab + b2<\/sup> = a2<\/sup> + b2<\/sup> + ab = 68 + 16 = 84<\/p>\n\n\n\n

Question 4: If a + b = 8 and ab = 6, find the value of a3<\/sup> + b3<\/sup>.<\/strong><\/p>\n\n\n\n

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

a + b = 8, ab = 6<\/p>\n\n\n\n

Cubing, a + b = 8, both sides, we get<\/p>\n\n\n\n

(a + b)3<\/sup> = (8)3<\/sup><\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 3ab(a + b) = 512<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 3 x 6 x 8 = 512<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> + 144 = 512<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> = 512 \u2013 144 = 368<\/p>\n\n\n\n

a3<\/sup> + b3<\/sup> = 368<\/p>\n\n\n\n

\n\n\n\n

Exercise 4.5 Page No: 4.28<\/h4>\n\n\n\n

Question 1: Find the following products:<\/strong><\/p>\n\n\n\n

(i) (3x + 2y + 2z) (9x2<\/sup> + 4y2<\/sup> + 4z2<\/sup> \u2013 6xy \u2013 4yz \u2013 6zx)<\/strong><\/p>\n\n\n\n

(ii) (4x \u2013 3y + 2z) (16x2<\/sup> + 9y2 <\/sup>+ 4z2 <\/sup>+ 12xy + 6yz \u2013 8zx)<\/strong><\/p>\n\n\n\n

(iii) (2a \u2013 3b \u2013 2c) (4a2<\/sup> + 9b2<\/sup> + 4c2<\/sup> + 6ab \u2013 6bc + 4ca)<\/strong><\/p>\n\n\n\n

(iv) (3x -4y + 5z) (9x2<\/sup> + 16y2<\/sup> + 25z2<\/sup> + 12xy- 15zx + 20yz)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> (3x + 2y + 2z) (9x2<\/sup> + 4y2<\/sup> + 4z2<\/sup> \u2013 6xy \u2013 4yz \u2013 6zx)<\/p>\n\n\n\n

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

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

= 27x3<\/sup> + 8y3<\/sup> + 8Z3<\/sup> \u2013 36xyz<\/p>\n\n\n\n

(ii)<\/strong> (4x \u2013 3y + 2z) (16x2<\/sup> + 9y2<\/sup> + 4z2<\/sup> + 12xy + 6yz \u2013 8zx)<\/p>\n\n\n\n

= (4x -3y + 2z) [(4x)2<\/sup> + (-3y) 2<\/sup> + (2z) 2<\/sup> \u2013 4x x (-3y) \u2013 (-3y) x (2z) \u2013 (2z x 4x)]<\/p>\n\n\n\n

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

= 64x3<\/sup> \u2013 27y3<\/sup> + 8z3<\/sup> + 72xyz<\/p>\n\n\n\n

(iii)<\/strong> (2a -3b- 2c) (4a2<\/sup> + 9b2<\/sup> + 4c2<\/sup> + 6ab \u2013 6bc + 4ca)<\/p>\n\n\n\n

= (2a -3b- 2c) [(2a) 2<\/sup> + (-3b) 2<\/sup> + (-2c) 2<\/sup> \u2013 2a x (-3b) \u2013 (-3b) x (-2c) \u2013 (-2c) x 2a]<\/p>\n\n\n\n

= (2a)3<\/sup> + (-3b) 3<\/sup> + (-2c) 3<\/sup> -3x 2a x (-3 b) (-2c)<\/p>\n\n\n\n

= 8a3<\/sup> \u2013 21b3<\/sup> \u2013 8c3<\/sup> \u2013 36abc<\/p>\n\n\n\n

(iv)<\/strong> (3x \u2013 4y + 5z) (9x2<\/sup> + 16y2<\/sup> + 25z2<\/sup> + 12xy \u2013 15zx + 20yz)<\/p>\n\n\n\n

= [3x + (-4y) + 5z] [(3x) 2<\/sup> + (-4y) 2<\/sup> + (5z) 2<\/sup> \u2013 3x x (-4y) -(-4y) (5z) \u2013 5z x 3x]<\/p>\n\n\n\n

= (3x) 3<\/sup> + (-4y) 3<\/sup> + (5z) 3<\/sup> \u2013 3 x 3x x (-4y) (5z)<\/p>\n\n\n\n

= 27x3<\/sup> \u2013 64y3<\/sup> + 125z3<\/sup> + 180xyz<\/p>\n\n\n\n

Question 2: If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3<\/sup> + y3<\/sup> + z3<\/sup> \u2013 3xyz.<\/strong><\/p>\n\n\n\n

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

We know, x3<\/sup> + y3<\/sup> + z3<\/sup> \u2013 3xyz = (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> \u2013 xy \u2013 yz \u2013 zx)<\/p>\n\n\n\n

Squaring, x + y + z = 8 both sides, we get<\/p>\n\n\n\n

(x + y + z)2<\/sup> = (8) 2<\/sup><\/p>\n\n\n\n

x2<\/sup> + y2<\/sup> + z2<\/sup> + 2(xy + yz + zx) = 64<\/p>\n\n\n\n

x2<\/sup> + y2<\/sup> + z2<\/sup> + 2 x 20 = 64<\/p>\n\n\n\n

x2<\/sup> + y2<\/sup> + z2<\/sup> + 40 = 64<\/p>\n\n\n\n

x2<\/sup> + y2<\/sup> + z2<\/sup> = 24<\/p>\n\n\n\n

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

x3<\/sup> + y3<\/sup> + z3<\/sup> \u2013 3xyz = (x + y + z) [x2<\/sup> + y2<\/sup> + z2<\/sup> \u2013 (xy + yz + zx)]<\/p>\n\n\n\n

= 8(24 \u2013 20)<\/p>\n\n\n\n

= 8 x 4<\/p>\n\n\n\n

= 32<\/p>\n\n\n\n

\u21d2 x3<\/sup> + y3<\/sup> + z3<\/sup> \u2013 3xyz = 32<\/p>\n\n\n\n

Question 3: If a +b + c = 9 and ab + bc + ca = 26, find the value of a3<\/sup> + b3<\/sup> + c3<\/sup> \u2013 3abc.<\/strong><\/p>\n\n\n\n

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

a + b + c = 9, ab + bc + ca = 26<\/p>\n\n\n\n

Squaring, a + b + c = 9 both sides, we get<\/p>\n\n\n\n

(a + b + c)2<\/sup> = (9)2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + c2<\/sup> + 2 (ab + bc + ca) = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + c2<\/sup> + 2 x 26 = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + c2<\/sup> + 52 = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + c2<\/sup> = 29<\/p>\n\n\n\n

Now, a3<\/sup> + b3<\/sup> + c3<\/sup> \u2013 3abc = (a + b + c) [(a2<\/sup> + b2<\/sup> + c2<\/sup> \u2013 (ab + bc + ca)]<\/p>\n\n\n\n

= 9[29 \u2013 26]<\/p>\n\n\n\n

= 9 x 3<\/p>\n\n\n\n

= 27<\/p>\n\n\n\n

\u21d2 a3<\/sup> + b3<\/sup> + c3<\/sup> \u2013 3abc = 27<\/p>\n\n\n\n

\n\n\n\n

Exercise VSAQs Page No: 4.28<\/h4>\n\n\n\n

Question 1: If x + 1\/x = 3, then find the value of x2<\/sup> + 1\/x2<\/sup>.<\/strong><\/p>\n\n\n\n

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

x + 1\/x = 3<\/p>\n\n\n\n

Squaring both sides, we have<\/p>\n\n\n\n

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

x2<\/sup> + 1\/x2<\/sup> + 2 = 9<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 9 \u2013 2 = 7<\/p>\n\n\n\n

Question 2: If x + 1\/x = 3, then find the value of x^6 + 1\/x^6.<\/strong><\/p>\n\n\n\n

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

x + 1\/x = 3<\/p>\n\n\n\n

Squaring both sides, we have<\/p>\n\n\n\n

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

x2<\/sup> + 1\/x2<\/sup> + 2 = 9<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 9 \u2013 2 = 7<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 7 \u2026(1)<\/p>\n\n\n\n

Cubing equation (1) both sides,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: If a + b = 7 and ab = 12, find the value of a2<\/sup> + b2<\/sup>.<\/strong><\/p>\n\n\n\n

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

a + b = 7, ab = 12<\/p>\n\n\n\n

Squaring, a + b = 7, both sides,<\/p>\n\n\n\n

(a + b) 2<\/sup> = (7) 2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2ab = 49<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2 x 12 = 49<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 24 = 49<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 25<\/p>\n\n\n\n

Question 4: If a \u2013 b = 5 and ab = 12, find the value of a2<\/sup> + b2<\/sup>.<\/strong><\/p>\n\n\n\n

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

a \u2013 b = 5, ab = 12<\/p>\n\n\n\n

Squaring, a \u2013 b = 5, both sides,<\/p>\n\n\n\n

(a \u2013 b)2<\/sup> = (5)2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> \u2013 2ab = 25<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> \u2013 2 x 12 = 25<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> \u2013 24 = 25<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 49<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 4: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 4\u2013Algebraic Identities<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 4\u2013Algebraic Identities PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n