{"id":623842,"date":"2023-08-31T11:20:46","date_gmt":"2023-08-31T11:20:46","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623842"},"modified":"2023-08-31T11:20:51","modified_gmt":"2023-08-31T11:20:51","slug":"kc-sinha-exercise-4-5-mathematics-solution-class-9-chapter-4-algebraic-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-5-mathematics-solution-class-9-chapter-4-algebraic-identities\/","title":{"rendered":"KC Sinha: Exercise 4.5 – Mathematics Solution Class 9 Chapter 4 Algebraic identities"},"content":{"rendered":"\n\n\n\n\n

Exercise 4.5 <\/h2>\n\n\n\n

Question 1<\/h4>\n\n\n\n

a3<\/sup>+27b3<\/sup><\/strong>
Sol:
\u21d2(a)3<\/sup>+(3b)3<\/sup>
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2(a+3b)[a2<\/sup>-(a)(3b)+(3b)2<\/sup>]
\u21d2(a+3b)(a2<\/sup>-3ab+9b2<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Q2 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

x3<\/sup>+125<\/strong>
Sol:
\u21d2x3<\/sup>+53<\/sup>
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2(x+5)[x2<\/sup>-(x)(5)+52<\/sup>]
\u21d2(x+5)(x2<\/sup>-5x+25)<\/p>\n\n\n\n

\n\n\n\n

Q3 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

8a3<\/sup>+27b3<\/sup><\/strong>
Sol:
\u21d2(2a)3<\/sup>+(3b)3<\/sup>
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2(2a+3b)[(2a)2<\/sup>-(2a)(3b)+(3b)2<\/sup>]
\u21d2(2a+3b)(4a2<\/sup>+9b2<\/sup>-6ab)<\/p>\n\n\n\n

\n\n\n\n

Q4 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

x5<\/sup>+27x<\/strong>2<\/sup><\/strong>
[Hint: x5<\/sup>+27x2<\/sup>=x2<\/sup>(x3<\/sup>+27)]
Sol:
[Taking common x2<\/sup>]
\u21d2x2<\/sup>(x3<\/sup>+27)
\u21d2x2<\/sup>(x3<\/sup>+33<\/sup>)
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2x2<\/sup>(x+3)(x2<\/sup>-3x+32<\/sup>)
\u21d2x2<\/sup>(x+3)(x2<\/sup>-3x+9)<\/p>\n\n\n\n

\n\n\n\n

Q5 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

ab7<\/sup>+ba7<\/sup><\/strong>
[Hint:ab7<\/sup>+ba7<\/sup>=ab(a6<\/sup>+b6<\/sup>)=ab[(a2<\/sup>)3<\/sup>+(b2<\/sup>)3<\/sup>]]
Sol:
[Taking common ab]
\u21d2ab(b6<\/sup>+a6<\/sup>)
\u21d2ab[(b2<\/sup>)3<\/sup>+(a2<\/sup>)3<\/sup>]
\u21d2ab[(a2<\/sup>)3<\/sup>+(b2<\/sup>)3<\/sup>]
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2ab[(a2<\/sup>+b2<\/sup>)][(a2<\/sup>)2<\/sup>-(a2<\/sup>)(b2<\/sup>)+(b2<\/sup>)2<\/sup>]
\u21d2ab(a2<\/sup>+b2<\/sup>)(a4<\/sup>-a2<\/sup>b2<\/sup>+b4<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Q6 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

8(a+b)3<\/sup> + 27(b+c)3<\/sup><\/strong>
Sol:
[Hint: Given expression {2(a+b)}]3<\/sup>+{3(b+c)}3<\/sup>
={2(a+b)+3(b+c)}{4(a+b)2<\/sup>-6(a+b)(b+c)+9(b+c)2<\/sup>}
=(2a+5b+3c){(a+b)(4a+4b-6b-6c)+9(b2<\/sup>+c2<\/sup>+2bc)}
=(2a+5b+3c){(a+b)(4a-2b-6c)+9b2<\/sup>+9c2<\/sup>+18bc}
=(2a+5b+3c)(4a2<\/sup>+7b2<\/sup>+9c2<\/sup>+2ab+12bc-6ac)<\/p>\n\n\n\n

\n\n\n\n

Q7 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

$125x^3+\\dfrac{1}{8}$<\/strong><\/p>\n\n\n\n

Sol:<\/p>\n\n\n\n

\u21d2$5x^3+\\left(\\dfrac{1}{2}\\right)^3$<\/p>\n\n\n\n

Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2$\\left(5x+\\dfrac{1}{2}\\right)$$+(5x)^2-(5x)\\left(\\dfrac{1}{2}\\right)+\\left(\\dfrac{1}{2}\\right)^2$
\u21d2$\\left(5x+\\dfrac{1}{2}\\right)$$+\\left(25x^2-\\dfrac{5}{2}x+\\dfrac{1}{4}\\right)$<\/p>\n\n\n\n

\n\n\n\n

Q8 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

a3<\/sup>+b3<\/sup>+a+b<\/strong>
Sol:
\u21d2a3<\/sup>+b3<\/sup>+a+b
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2(a+b)(a2<\/sup>-ab+b2<\/sup>)+(a+b)
[Taking common (a+b)]
\u21d2(a+b)[(a2<\/sup>-ab+b2<\/sup>)+1]
\u21d2(a+b)(a2<\/sup>-ab+b2<\/sup>+1)<\/p>\n\n\n\n

\n\n\n\n

Q9 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

x3<\/sup>+y3<\/sup>+2x2<\/sup>-2y2<\/sup><\/strong>
Sol:
\u21d2(x3<\/sup>+y3<\/sup>)+2(x2<\/sup>-y2<\/sup>)
Using identity:
x3<\/sup>+y3<\/sup>=(x+y)(x2<\/sup>-xy+y2<\/sup>)
\u21d2[(x+y)(x2<\/sup>-xy+y2<\/sup>)]+2(x2<\/sup>-y2<\/sup>)
Using identity:
a2<\/sup>-b2<\/sup>=(a+b)(a-b)
\u21d2(x+y)(x2<\/sup>-xy+y2<\/sup>)+2(x+y)(x-y)
[Taking common (x+y)]
\u21d2(x+y)[(x2<\/sup>-xy+y2<\/sup>)+2(x-y)]
\u21d2(x+y)[x2<\/sup>-xy+y2<\/sup>+2x-2y]
\u21d2(x+y)(x2<\/sup>+y2<\/sup>-xy+2x-2y)<\/p>\n\n\n\n

\n\n\n\n

Q10 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

64x3<\/sup>-27y3<\/sup><\/strong>
Sol:
\u21d264x3<\/sup>-27y3<\/sup>
\u21d2(4x)3<\/sup>-(3y)3<\/sup>
Using identity:
x3<\/sup>-y3<\/sup>=(x-y)(x2<\/sup>+xy+y2<\/sup>)
\u21d2(4x-3y)[(4x)2<\/sup>-(4x)(3y)+(3y)2<\/sup>]
\u21d2(4x-3y)(16x2<\/sup>+12xy+9y2<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Q11 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

(a+b)3<\/sup>-(2)3<\/sup><\/strong>
[Hint: Given expression=(a+b-2)[(a+b)2<\/sup>+2(a+b)+4]]
Sol:
Using identity:
x3<\/sup>-y3<\/sup>=(x-y)(x2<\/sup>+xy+y2<\/sup>)
\u21d2(a+b-2)[(a+b)2<\/sup>+2(a+b)+22<\/sup>]<\/p>\n\n\n\n

$=(a+b-2)\\left(a^{2}+b^{2}+2 a b+2^{2}+2 a+2 b\\right]$<\/p>\n\n\n\n

$=(a+b-2)\\left(a^{2}+b^{2}+4+2 a+2 b\\right)$<\/p>\n\n\n\n

\n\n\n\n

Q12 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

x3<\/sup>y3<\/sup>-512<\/strong>
[Hint: Given expression =(xy)3<\/sup>-(8)3<\/sup> ]
Sol :
\u21d2(xy)3<\/sup>-83<\/sup>
Using identity:
x3<\/sup>-y3<\/sup>=(x-y)(x2<\/sup>+xy+y2<\/sup>)
\u21d2(xy-8)[(xy)2<\/sup>+(8)(xy)+(8)2<\/sup>]
\u21d2(xy-8)(x2<\/sup>y2<\/sup>+8xy+64)<\/p>\n\n\n\n

\n\n\n\n

Q13 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

$a^3-\\dfrac{27}{a^3}$
Sol :
Expression$=a^3-\\left(\\dfrac{3}{a}\\right)^3$
Using identity:
x3<\/sup>-y3<\/sup>=(x-y)(x2<\/sup>+xy+y2<\/sup>)
$=\\left(a-\\dfrac{3}{a}\\right)\\left(a^2+a.\\dfrac{3}{a}+\\dfrac{9}{a^2}\\right)$
\u21d2$\\left(a-\\dfrac{3}{a}\\right)$$+\\left(a^2+a\\dfrac{3}{a}+\\dfrac{9}{a^2}\\right)$<\/p>\n\n\n\n

\n\n\n\n

Page 4.35<\/p>\n\n\n\n

\n\n\n\n

Type 2
Problems based on the identity:<\/strong>
a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc=(a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>-ab-bc-ca)
Category A: <\/strong>Problems based on factorization of thepolynomials of the form a3<\/sup>+b3<\/sup>+c3 <\/sup>, whena+b+c=0
WORKING RULE:<\/strong>
Find the algebric sum of the terms. If it is zero, then required factors willbe 3\u00d7product of three terms.
Thus if a+b+c=0 , then a3<\/sup>+b3<\/sup>+c3<\/sup>=3abc<\/p>\n\n\n\n

Category B:<\/strong> Problems based on factorization of thepolynomials of the forma3<\/sup>+b3<\/sup>+c3 <\/sup>-3abc
WORKING RULE:<\/strong>
1. If polynomial is of the forma3<\/sup>+b3<\/sup>+c3 <\/sup>-3abc , then use identitya3<\/sup>+b3<\/sup>+c3<\/sup>-3abc=(a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>-ab-bc-ca)
2. If polynomial can be converted in the forma3<\/sup>+b3<\/sup>+c3 <\/sup>-3abc , then convert the givenexpression in this form and then factorize.
3. Use the following formulae whichever is required :
(i)a2<\/sup>+b2<\/sup>+c2<\/sup>-ab-bc-ca=1\/2{(a-b)2<\/sup>+(b-c)2<\/sup>+(c-a)2<\/sup>}
(ii)(a+b+c)3<\/sup>=a3<\/sup>+b3<\/sup>+c3<\/sup>+3(a+b)(b+c)(c+a)
(iii)a3<\/sup>+b3<\/sup>+c3<\/sup>=(a+b+c)3<\/sup>-3(a+b)(b+c)(c+a)
(iv)(a+b+c)3<\/sup>-a3<\/sup>-b3<\/sup>-c3<\/sup>=3(a+b)(b+c)(c+a)<\/p>\n\n\n\n

\n\n\n\n

Q14 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

Fill up the blanks.<\/strong><\/p>\n\n\n\n

\n\n\n\n

(i) If x+y+z=0 , factors of x3<\/sup>+y3<\/sup>+z3<\/sup> willbe __<\/strong>
Sol :
Using identity :
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>-3xyz=(x+y+z)(x2<\/sup>+y2<\/sup>+z2<\/sup>-xy-yz-zx)
[Given x+y+z=0]
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>-3xyz=(0<\/p>\n\n\n\n

)(x2<\/sup>+y2<\/sup>+z2<\/sup>-xy-yz-zx)
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>-3xyz=0<\/p>\n\n\n\n

\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>=3xyz<\/p>\n\n\n\n

\u2234 Factors of x3<\/sup>+y3<\/sup>+z3 <\/sup>is
\u21d23xyz<\/p>\n\n\n\n

\n\n\n\n

(ii) Factors of (a-b)3<\/sup>+(b-c)3<\/sup>+(c-a)3<\/sup> willbe __<\/strong>
Sol :
\u21d2(a-b)3<\/sup>+(b-c)3<\/sup>+(c-a)3<\/sup>
Using identity :
\u21d2a3<\/sup>+b3<\/sup>+c3<\/sup>=(a+b+c)3<\/sup>-3(a+b)(b+c)(c+a)
where a=(a-b) , b=(b-c) , c=(c-a)
\u21d2[(a-b)+(b-c)+(c-a)]-3[(a-b)+(b-c)][(b-c)+(c-a)][(c-a)+(a-b)]
\u21d2[a-b+b-c+c-a]-3[a-b+b-c][b-c+c-a][c-a+a-b]
\u21d2[0]-3[a-c][b-a][c-b]
\u21d23(a-b)(c-a)(b-c)
\u21d23(a-b)(b-c)(c-a)<\/p>\n\n\n\n

\n\n\n\n

(iii) Factors of (2a-3b)3<\/sup>+(3b-c)3<\/sup>+(c-2a)3<\/sup>will be __<\/strong>
Sol :
$=[2 a-3 b+3 b-c+c-2 a]^{3}-3(2 a-3 b+3 b-c)(3b-c+c-2a)(c-2a+2a-3b)$<\/p>\n\n\n\n

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

=-3\u00d7-1\u00d7-1\u00d7-1\u00d7(2a-3b)(3b-c)(c-2a)<\/p>\n\n\n\n

\u21d23(2a-3b)(3b-c)(c-2a)<\/p>\n\n\n\n

\n\n\n\n

(iv) Factors of (a+b+c)3<\/sup>+(a-b-c)3<\/sup>-8a3<\/sup> willbe __<\/strong>
[Hint: Given expression=(a+b+c)3<\/sup>+(a-b-c)3<\/sup>+(-2a)3<\/sup>]
Let x=a+b+c , y=a-b-c and z=-2a
Then , x+y+z=(a+b+c)+(a-b-c)-2a=0
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>=3xyz
\u21d2(a+b+c)3<\/sup>+(a-b-c)3<\/sup>-8a3<\/sup>
\u21d23(a+b+c)(a-b-c)(-2a)
\u21d2-6a(a+b+c)(a-b-c)
\u21d26a(a+b+c)(b+c-a)<\/p>\n\n\n\n

\n\n\n\n

(v) If p=2-a, prove that a3<\/sup>+p3<\/sup>-8+6ap=0<\/strong><\/p>\n\n\n\n

[Hint: We have, p=2-a]
\u21d2a+p-2=0
Let x=a , y=p and z=(-2)
Then, x+y+z=0
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>=3xyz
\u21d2(a)3<\/sup>+(p)3<\/sup>+(-2)3<\/sup>=3\u00d7a\u00d7p\u00d7(-2)
\u21d2a3<\/sup>+p3<\/sup>-8=-6ap or
a3<\/sup>+p3<\/sup>+6ap-8=0<\/p>\n\n\n\n

Second method:<\/strong>
=a3<\/sup>+6ap+p3<\/sup>-8=a3<\/sup>+p3<\/sup>+(-2)3<\/sup>-3ap(-2)
={a+p+(-2)}{a2<\/sup>+p2<\/sup>+(-2)2<\/sup>-ap-p(-2)-a(-2)}
=(a+p-2)(a2<\/sup>+p2<\/sup>+4-ap+2p+2a)
=0\u00d7(a2+p2+4-ap+2p+2a)=0<\/p>\n\n\n\n

\n\n\n\n

Q15 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

Factorize the following :<\/strong>
(i) x3<\/sup>-y3<\/sup>-z3<\/sup>-3xyz<\/strong>
Sol :<\/p>\n\n\n\n

$=(x-y-z)\\left(x^{2}+(-y)^{2}+(-z)^{2}-(x)(-y)-(-y)(-z)-(-z)(x)\\right.$<\/p>\n\n\n\n

\u21d2(x-y-z)(x2<\/sup>+y2<\/sup>+z2<\/sup>+xy+xz-yz)<\/p>\n\n\n\n

$=a^{3}-b^{3}-1^{3}-3 a b$<\/p>\n\n\n\n

$[a+(-b)+(-1)]\\left[a^{2}+(-b)^{2}+(-1)^{2}-(a)(-b)-(-b)(-1)-(-1)(a)\\right.$
\u21d2(a-b-1)(a2<\/sup>+b2<\/sup>+1+ab+a-b)<\/p>\n\n\n\n

$=(a)^{3}+(b)^{3}+(-c)^{3}-3(a)(b)-(c)$<\/p>\n\n\n\n

$=(a+b-c)\\left[a^{2}+b^{2}+(-c)^{2}-(a)(b)-(b)(-c)-(-c)(a)\\right]$
\u21d2(a+b-c)(a2<\/sup>+b2<\/sup>+c2<\/sup>-ab+bc+ca)<\/p>\n\n\n\n

\n\n\n\n

(iv) x3<\/sup>+8y3<\/sup>-z3<\/sup>+6xyz<\/strong>
Sol :
$\\Rightarrow(x)^{3}+2^{3} \\cdot y^{3}+(-z)^{3}-3(x)(2 y)(-z)$<\/p>\n\n\n\n

$\\Rightarrow(x)^{3}+(2 y)^{3}+(-z)^{3}-3(x)(+2 y)(-z)$<\/p>\n\n\n\n

$\\Rightarrow(x+2 y-z)\\left(x^{2}+(2 y)^{2}+(-z)^{2}-(x)(2 y)\\right.-(2y)(-z)-(-z)(x))$<\/p>\n\n\n\n

\u21d2(x+2y-z)(x2<\/sup>+4y2<\/sup>+z2<\/sup>-2xy+xz+2yz)<\/p>\n\n\n\n

\n\n\n\n

Q16 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

Find value of x3<\/sup>-8y3<\/sup>-36xy-216, if x=2y+6<\/strong>
[Hint: Given, x-2y-6=0
Let a=x , b=-2y and c=-6
Then , a+b+c=0
\u21d2 a3<\/sup>+b3<\/sup>+c3<\/sup>=3abc
\u21d2x3<\/sup>+(-2y)3<\/sup>+(-6)3<\/sup>=3x(-2y)(-6)
\u21d2x3<\/sup>-8y3<\/sup>-216=36xy
\u21d2x3<\/sup>-8y3<\/sup>-36xy-216=0<\/p>\n\n\n\n

Page 4.36
Second method :<\/strong>
Given , x-2y=6
\u21d2(x-2y)3<\/sup>=(6)3<\/sup>
\u21d2x3<\/sup>-3x2y(x-2y)-8y3<\/sup>=216
\u21d2x3<\/sup>-6xy(6)-8y3<\/sup>-216=0<\/p>\n\n\n\n

\u21d2x3<\/sup>-36xy-8y3<\/sup>-216=0<\/p>\n\n\n\n

\n\n\n\n

Q17 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

Find the value of a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc,if<\/strong>
(i) a+b+c=14 and a2<\/sup>+b2<\/sup>+c2<\/sup>=68<\/strong>
[Hint: (i) a+b+c=14
\u21d2(a+b+c)2<\/sup>=196
\u21d2a2<\/sup>+b2<\/sup>+c2<\/sup>+2(ab+bc+ac)=196
\u21d268+2(ab+bc+ac)=196
\u21d2ab+bc+ac$=\\dfrac{196-68}{2}=64$
Now ,a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc=(a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>-ab-bc-ca)
=14\u00d7(68-64)=56]
Sol :<\/p>\n\n\n\n

\u21d256<\/p>\n\n\n\n

\n\n\n\n

(ii) a+b+c=9 and ab+bc+ca=26<\/strong>
Sol:<\/p>\n\n\n\n

$a^{3}+b^{3}+c^{3}-3 abc=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)$<\/p>\n\n\n\n

$=9\\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\\right)$<\/p>\n\n\n\n

$=9\\left(a^{2}+b^{2}+c^{2}-26\\right)$…(i)<\/p>\n\n\n\n

ab+bc+ca=26<\/p>\n\n\n\n

Taking square both sides<\/p>\n\n\n\n

$(a+b+c)^{2}=81$<\/p>\n\n\n\n

$a^{2}+b^{2}+c^{2}+2(a b+b c+ca)=81$<\/p>\n\n\n\n

$a^{2}+b^{2}+c^{2}+2 \\times 26=81$<\/p>\n\n\n\n

$a^{2}+b^{2}+c^{2}=81-52=29$…(ii)<\/p>\n\n\n\n

Putting (ii) in (i)<\/p>\n\n\n\n

=9(29-26)=9\u00d73=27<\/p>\n\n\n\n

\n\n\n\n

Q18 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 18<\/h4>\n\n\n\n

Find the value of x3<\/sup>+y3<\/sup>+z3<\/sup> , if x+y+z=11,<\/strong>
x2<\/sup>+y2<\/sup>+z2<\/sup>=45 and xyz=40<\/strong>
Sol :
$x^{3}+y^{3}+z^{3}-3xy z=(x+y+z)\\left(x^{2}-y^{2}+z^{2}-x y-y z-zx\\right)$<\/p>\n\n\n\n

$x^{3}+y^{3}+z^{3}-3 \\times 40=(11)(45-(xy+yz+zx)$…(i)<\/p>\n\n\n\n

x+y+z=11<\/p>\n\n\n\n

$(x+y+z)^{2}=11^{2}$<\/p>\n\n\n\n

$x^{2}+y^{2}+z^{2}+2\\left(x y+yz+zx\\right)=121$<\/p>\n\n\n\n

$45+2(x y+y z+2 x)=121$<\/p>\n\n\n\n

xy+yz+zx$=\\frac{121-45}{2}=38$…(ii)<\/p>\n\n\n\n

Putting (ii) in (i)<\/p>\n\n\n\n

$x^{3}+y^{3}+z^{3}-120=11(45-38)$<\/p>\n\n\n\n

$x^{3}+y^{3}+z^{3}-120=11 \\times 7$<\/p>\n\n\n\n

$x^{3}+y^{3}+z^{3}=77+120=197$<\/p>\n\n\n\n

\n\n\n\n

Q19 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

Find the product<\/strong>
(i) (x+y-z)(x2<\/sup>+y2<\/sup>+z2<\/sup>-xy+yz+zx)<\/strong>
Sol :<\/p>\n\n\n\n

$=(x+y-z)\\left(x^{2}+y^2+z^{2}-xy+y z+z x\\right)$<\/p>\n\n\n\n

$=[x+y+(-z)]\\left[x^{2}+y^{2}+(-z)^{2}-(x) (y\\right)-(y)(-z)-(-z)(x)]$<\/p>\n\n\n\n

$=x^{3}+y^{2}+(-z)^{3}-3(x)(y)(-z)$
\u21d2(x3<\/sup>+y3<\/sup>-z3<\/sup>+3xyz)<\/p>\n\n\n\n

\n\n\n\n

(ii) (3x-5y-4)(9x2<\/sup>+25y2<\/sup>+15xy+12x-20y+16)<\/strong>
[Hint: (i) We have(x+y-z)(x2<\/sup>+y2<\/sup>+z2<\/sup>-xy+yz+zx)]
=[x+y+(-z)][x2<\/sup>+y2<\/sup>+(-z)2<\/sup>-xy-y(-z)-x(-z)]
=(a+b+c)(a2<\/sup>+b2<\/sup>+c2<\/sup>-ab-bc-ac) , where a=x , b=yand c=-z
=a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc=27x3<\/sup>-25y3<\/sup>-64-3(3x)(-5y)(-4)
=27x3<\/sup>-125y3<\/sup>-64-180xy<\/p>\n\n\n\n

\n\n\n\n

Q20 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

(i) Find the value of x3<\/sup>+y3<\/sup>+z3<\/sup> if x+y+z=1,<\/strong>
xy+yz+zx=-1 and xyz=-1<\/strong>
[Hint: (i) We know thatx3<\/sup>+y3<\/sup>+z3<\/sup>-3xyz
=(x+y+z)(x2<\/sup>+y2<\/sup>+z2<\/sup>-xy-yz-zx)]
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>-3xyz=(x+y+z)(x2<\/sup>+y2<\/sup>+z2<\/sup>+2xy+2yz+2zx-3xy-3yz-3zx)
[Addingand subtracting 2xy+2yz+2zx in second bracket only]
\u21d2x3<\/sup>+y3<\/sup>+z3<\/sup>=(x+y+z)(x+y+z)2-3(xy+yz+zx)]+3xyz
[Transposing3xyz on R.H.S]
=1\u00d7[(1)2<\/sup>-3(-1)]+3(-1)
= 4-3 = 1<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u21d21<\/p>\n\n\n\n

\n\n\n\n

(ii) Simplify <\/strong>$\\frac{\\left(a^{2}-b^{2}\\right)^{3}+\\left(b^{2}-c^{2}\\right)^{3}+\\left(c^{2}-a^{2}\\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

(a2<\/sup>-b2<\/sup>)+(b2<\/sup>-c2<\/sup>)+(c2<\/sup>-a2<\/sup>)=0<\/p>\n\n\n\n

\u2234Numerator
=(a2<\/sup>-b2<\/sup>)+(b2<\/sup>-c2<\/sup>)+(c2<\/sup>-a2<\/sup>)=3(a2<\/sup>-b2<\/sup>)(b2<\/sup>-c2<\/sup>)(c2<\/sup>-a2<\/sup>)
=3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)..(i)
Again, (a-b)+(b-c)+(c-a)=0
\u2234(a-b)3<\/sup>+(b-c)3<\/sup>+(c-a)3<\/sup>=3(a-b)(b-c)(c-a)..(ii)
Hence, given expression=(a+b)(b+c)(c+a)<\/p>\n\n\n\n

\n\n\n\n

Q21 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper<\/strong><\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

Prove that:<\/strong>
(i) (a3<\/sup>+b3<\/sup>+c3<\/sup>)-3abc=1\/2(a+b+c){(a-b)2<\/sup>+(b-c)2<\/sup>+(c-a)2<\/sup>}<\/strong>
Sol :<\/p>\n\n\n\n

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\\right)$<\/p>\n\n\n\n

$=\\frac{1}{2} \\times(a+b+c) \\left\\{a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 bc\\right.\\left.+c^{2}+a^{2}-2 c a\\right\\}$<\/p>\n\n\n\n

$=\\frac{1}{2} \\times (a+b+c) \\left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b\\right. -2 bc-2 c a)$<\/p>\n\n\n\n

$=2\\times \\frac{1}{2}(a+b+c)\\left\\{\\left(a^{2}+b^{2}+c^{2}-a b-b c-ca\\right\\}\\right.$<\/p>\n\n\n\n

$a^{2}+b^{3}+c^{3}-3 a b c$<\/p>\n\n\n\n

\n\n\n\n

(ii) (a+b)3<\/sup>+(b+c)3<\/sup>+(c+a)3<\/sup>-3(a+b)(b+c)(c+a)=2(a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc)<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\Rightarrow(a+b)^{3}+(b+c)^{3}+(c+a)^{3}-3(a+b)(b+c)(c+a)$<\/p>\n\n\n\n

$\\left[(a+b)+(b+c)+(c+a)]\\left((a+b)^{2}+(b+c)^{2}+\\left(c+a)^{2}\\right.\\right.\\right.-(a+b)(b+c)-(b+c)(c+a)-((c+a)(a+b)\\}$<\/p>\n\n\n\n

$\\Rightarrow(2 a+2 b+2 c)\\left\\{a^{2}+b^{2}+2 a b+b^{2}+c^{2}+2 bc+c+a^{2}\\right.+2 c a-[a(b+c)+b(b+c)]-[b(c+a)+c(c+a)]-[c(a+b)+a(a+b)]$<\/p>\n\n\n\n

$=2(a+b+c) \\{2 a^{2}+2 b^{2}+2 c^{2}+2 a b+2 b c+2 c a-\\left[a b+a c+b^{2}+bc\\right]-\\left[b c+b a+c^{2}+c a\\right]-\\left[c a+c b+a^{2}\\right.+ab]$<\/p>\n\n\n\n

$\\Rightarrow 2(a+b+c)\\left[2 a^{2}+2 b^{2}+2 c^{2}+2 a b+2 b c+2 c a-a b-ac-b^{2}-b c\\right. \\left.-b c-b a-c^{2}-c a-c a-c b-a^{2}-a b\\right\\}$<\/p>\n\n\n\n

$\\Rightarrow 2(a+b+c)\\left\\{a^{2}+b^{2}+c^{2}+a b+b c+c a-a c-b c-b a-ca-cb-ab\\right\\}$<\/p>\n\n\n\n

$\\Rightarrow \\quad 2(a+b+c)\\left(a^{2}+b^{2}+c^{2}-c a-c b-a b\\right)$<\/p>\n\n\n\n

 =2(a3<\/sup>+b3<\/sup>+c3<\/sup>-3abc)<\/p>\n\n\n\n

(iii) Factorize :p3<\/sup>(q-r)3<\/sup>+q3<\/sup>(r-p)3<\/sup>+r3<\/sup>(p-q)3<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\Rightarrow[p(q-r)]^{3}+[q(r-p)]^{3}+[r(p-q)]^{3}$<\/p>\n\n\n\n

[p(q-r)]=x , [q(r-p)]=y , [r(p-q)]=z<\/p>\n\n\n\n

x3<\/sup>+y3<\/sup>+z3<\/sup>=3xyz, when x+y+z=0<\/p>\n\n\n\n

=[p(q-r)]+[q(r-p)]+[r(p-q)]<\/p>\n\n\n\n

=pq-rp+rq-qp+rp-rq<\/p>\n\n\n\n

=0<\/p>\n\n\n\n

\u22343[p(q-r)]+[q(r-p)]+[r(p-q)]<\/p>\n\n\n\n

\n\n\n\n

(iv) Find the value of(x-a)3<\/sup>+(x-b)3<\/sup>+(x-c)3<\/sup>=3(x-a)(x-b)(x-c) whena+b+c=3x<\/strong>
Sol :<\/p>\n\n\n\n

$(x-a)^{3}+(x-b)^{3}+(x-c)^{3}=3(x-a)(x-b)(x-c)$<\/p>\n\n\n\n

$(x-a)^{3}+(x-b)^{3}+(x-c)^{3}-3(x-a)(x-b)(x-c)$<\/p>\n\n\n\n

Using identity<\/p>\n\n\n\n

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\\right)$<\/p>\n\n\n\n

$=[(x-a)+(x-b)+(x-c)]\\left[(x-a)^{2}+(x-b)^{2}+(x-c)^{2}-(x-a)(x-b)-(x-b)(x-c)-(x-c)(x-a)\\right]$<\/p>\n\n\n\n

$=[x-a+x-b+x-c]\\left[x^{2}+a^{2}-2 a x+x^{2}+b^{2}-2 x b+x^{2}+c^{2}\\right.-2x c-[x(x-b)-a(x-b)]-[x(x-c)-b(x-c)]-[x(x-a)-c(x-a)]$<\/p>\n\n\n\n

$\\Rightarrow[3x-a-b-c]\\left\\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\\right\\}$<\/p>\n\n\n\n

$\\Rightarrow[3x-(a+b+c)]\\left\\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\\right\\}$<\/p>\n\n\n\n

 \u2235a+b+c=3x<\/p>\n\n\n\n

$\\Rightarrow[3x-(3x)]\\left\\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\\right\\}$<\/p>\n\n\n\n

$\\Rightarrow[0]\\times\\left\\{3x^{2}+a^{2}-2xa+b^{2}-2xb+c^{2}-2xc-[x^2-xb-xa+ab]-[x^2-xc-bx+bc]-[x^2-xa-cx+ca]\\right\\}$<\/p>\n\n\n\n

=0<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Exercise 4.5 Question 1 a3+27b3Sol:\u21d2(a)3+(3b)3Using identity:x3+y3=(x+y)(x2-xy+y2)\u21d2(a+3b)[a2-(a)(3b)+(3b)2]\u21d2(a+3b)(a2-3ab+9b2) Q2 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper Question 2 x3+125Sol:\u21d2x3+53Using identity:x3+y3=(x+y)(x2-xy+y2)\u21d2(x+5)[x2-(x)(5)+52]\u21d2(x+5)(x2-5x+25) Q3 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper Question 3 8a3+27b3Sol:\u21d2(2a)3+(3b)3Using identity:x3+y3=(x+y)(x2-xy+y2)\u21d2(2a+3b)[(2a)2-(2a)(3b)+(3b)2]\u21d2(2a+3b)(4a2+9b2-6ab) Q4 | Ex-4.5 | Class 9 |Algebraic Identities | KC SINHA Mathematics |myhelper Question 4 […]<\/p>\n","protected":false},"author":302,"featured_media":623835,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[921],"tags":[],"boards":[],"class_list":["post-623842","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 4.5 - 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