RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities - Free PDF Download

Hey, are you a class 9 student and looking for ways to download RD Sharma Solutions for Class 9 Maths Chapter 4  "Algebraic Identities"? If yes. Then read this post till the end.

In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 4 in PDF that is prepared by Kota’s top IITian’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 9 Maths Chapter 4 "Algebraic Identities" in an easy way then you can use these solutions PDF.

Chapter 4 of RD Sharma Class 9 deals with Algebraic identities, they are actually algebraic equations which are always true for every value of variables in them. If you want to improve your basic fundamentals of Algebraic Identities then you can use this.

RD Sharma Solutions helps students to Practice important concepts of subjects easily. RD Sharma class 9 solutions provide detailed explanations of all the  exercise questions that students can use to clear their doubts instantly.

If you want to become good at Math then it is very important for you to have a good knowledge of all the important topics of class 9 math, so to learn and practice those topics you can use eSaral RD Sharma Solutions.

In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 4 that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities - Free PDF Download

Question 1: Evaluate each of the following using identities:

(i) $(2 x-1 / x)^{2}$

(ii) $(2 x+y)(2 x-y)$

(iii) $\left(a^{2} b-b^{2} a\right)^{2}$

(iv) $(a-0.1)(a+0.1)$

(v) $\left(1.5 . x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)$

Solution: (i) $(2 x-1 / x)^{2}$

[Use identity: $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$ $(2 x-1 / x)^{2}=(2 x)^{2}+(1 / x)^{2}-2(2 x)(1 / x)$

$=4 x^{2}+1 / x^{2}-4$

(ii) $(2 x+y)(2 x-y)$

[Use identity: $\left.(a-b)(a+b)=a^{2}-b^{2}\right]$

$(2 x-1 / x)^{2}=(2 x)^{2}+(1 / x)^{2}-2(2 x)(1 / x)$

$=4 x^{2}-y^{2}$

(iii) $\left(\mathrm{a}^{2} \mathrm{~b}-\mathrm{b}^{2} \mathrm{a}\right)^{2}$

[Use identity: $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$\left(a^{2} b-b^{2} a\right)^{2}=\left(a^{2} b\right)^{2}+\left(b^{2} a\right)^{2}-2\left(a^{2} b\right)\left(b^{2} a\right)$

$=a^{4} b^{2}+b^{4} a^{2}-2 a^{3} b^{3}$

(iv) $(a-0.1)(a+0.1)$

[Use identity: $\left.(a-b)(a+b)=a^{2}-b^{2}\right]$

$(a-0.1)(a+0.1)=(a)^{2}-(0.1)^{2}$

$=(a)^{2}-0.01$

(v) $\left(1.5 x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)$

[Use identity: $\left.(a-b)(a+b)=a^{2}-b^{2}\right]$

$\left(1.5 x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)$

$=\left(1.5 x^{2}\right)^{2}-\left(0.3 y^{2}\right)^{2}$

$=2.25 x^{4}-0.09 y^{4}$

Question 2: Evaluate each of the following using identities:

(i) $(399)^{2}$

(ii) $(0.98)^{2}$

(iii) $991 \times 1009$

(iv) $117 \times 83$

Solution: (i) $399^{2}=(400-1)^{2}$

$=(400)^{2}+(1)^{2}-2 \times 400 \times 1$

[Use identity: $(\mathrm{a}-\mathrm{b})^{2}$

$\left.=a^{2}+b^{2}-2 a b\right]$

Here, $a=400$ and $b=1$

$=160000+1-8000$

$=159201$

So, $(399)^{2}=159201$

(ii) $(0.98)^{2}=(1-0.02)^{2}$

[Use identity: $(\mathrm{a}-\mathrm{b})^{2}$

$\left.=a^{2}+b^{2}-2 a b\right]$

$=(1)^{2}+(0.02)^{2}-2 \times 1 \times 0.02$

$=1+0.0004-0.04$

$=1.0004-0.04$

$=0.9604$

So, $(0.98)^{2}=0.9604$

(iii) $991 \times 1009$

$=(1000-9)(1000+9)$

[Use identity: (a - b) (a + b)

$\left.=a^{2}-b^{2}\right]$

$=(1000)^{2}-(9)^{2}$

$=1000000-81$

$=999919$

$991 \times 1009=999919$

(iv) $117 \times 83$

$=(100+17)(100-17)$

[Use identity: $(a-b)(a+b)$

$\left.=a^{2}-b^{2}\right]$

$=(100)^{2}-(17)^{2}$

$=10000-289$

$=9711$

$117 \times 83=9711$

Question 3: Simplify each of the following:

(i) $175 \times 175+2 \times 175 \times 25+25 \times 25$

(ii) $322 \times 322-2 \times 322 \times 22+22 \times 22$

(iii) $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$

(iv) $\frac{7.83 \times 7.83-1.17 \times 1.17}{6.66}$

Solution: (i) $175 \times 175+2 \times 175 \times 25+25 \times 25$

$=(175)^{2}+2(175)(25)+(25)^{2}$

$=(175+25)^{2}$

[Because $\left.a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$

$=(200)^{2}$

$=40000$

So, $175 \times 175+2 \times 175 \times 25+25 \times 25$

$=40000$

(ii) $322 \times 322-2 \times 322 \times 22+22 \times 22$

$=(322)^{2}-2 \times 322 \times 22+(22)^{2}$

$=(322-22)^{2}$

[Because $\left.a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$

$=(300)^{2}$

$=90000$

So, $322 \times 322-2 \times 322 \times 22+22 \times 22$

$=90000$

(iii) $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$

$=(0.76)^{2}+2 \times 0.76 \times 0.24+(0.24)^{2}$

$=(0.76+0.24)^{2}$

[ Because $\left.a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$

$=(1.00)^{2}$

$=1$

So, $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24=1$

(iv) $\frac{7.83 \times 7.83-1.17 \times 1.17}{6.66}$

$=\frac{(7.83+1.17)(7.83-1.17)}{6.66}$

$=\frac{(9.00)(6.66)}{(6.66)}=9$

Question 4: If $x+1 / x=11$, find the value of $x^{2}+1 / x^{2}$.

Solution: $x+\frac{1}{x}=11$ (Given)

So, $\left(x+\frac{1}{x}\right)^{2}$

$x^{2}+\left(\frac{1}{x}\right)^{2}+2 \times x \times \frac{1}{x}$

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}$

$=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow(11)^{2}$

$=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow 121$

$=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}$

$=119$

Question 5: If $x-1 / x=-1$, find the value of $x^{2}+1 / x^{2}$

Solution: $x-\frac{1}{x}=-1 \text { (Given) }$

So, $\left(x-\frac{1}{x}\right)^{2}$

$x^{2}+\left(\frac{1}{x}\right)^{2}-2 \times x \times \frac{1}{x}$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}$

$=x^{2}+\frac{1}{x^{2}}-2$

$\Rightarrow(-1)^{2}$

$=x^{2}+\frac{1}{x^{2}}-2$

$\Rightarrow 2+1$

$=x^{2}+\frac{1}{x^{2}}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}$

$=3$

Exercise $4.2$ Page No: $4.11$

Question 1: Write the following in the expanded form:

(i) $(a+2 b+c)^{2}$

(ii) $(2 a-3 b-c)^{2}$

(iii) $(-3 x+y+z)^{2}$

(iv) $(m+2 n-5 p)^{2}$

$(v)(2+x-2 y)^{2}$ (vi) $\left(\mathbf{a}^{2}+\mathbf{b}^{2}+\mathbf{c}^{2}\right)^{2}$

$(v i i)(a b+b c+c a)^{2}$

(viii) $(x / y+y / z+z / x) 2$

$(\mathrm{ix})(\mathrm{a} / \mathrm{bc}+\mathrm{b} / \mathrm{ac}+\mathrm{c} / \mathrm{ab})^{2}$

$(x)(x+2 y+4 z)^{2}$

$(x i)(2 x-y+z)^{2}$

$(x i i)(-2 x+3 y+2 z)^{2}$

Solution: Using identities:

$(x+y+z)^{2}$

$=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z$

(i) $(a+2 b+c)^{2}$

$=a^{2}+(2 b)^{2}+c^{2}+2 a(2 b)+2 a c+2(2 b) c$

$=a^{2}+4 b^{2}+c^{2}+4 a b+2 a c+4 b c$

(ii) $(2 a-3 b-c)^{2}$

$=[(2 a)+(-3 b)+(-c)]^{2}$

$=(2 a)^{2}+(-3 b)^{2}+(-c)^{2}+2(2 a)(-3 b)$

$+2(-3 b)(-c)+2(2 a)(-c)$

$=4 a^{2}+9 b^{2}+c^{2}-12 a b+6 b c-4 c a$

(iii) $(-3 x+y+z)^{2}$

$=\left[(-3 x)^{2}+y^{2}+z^{2}+2(-3 x) y+2 y z+2(-3 x) z\right.$

$=9 x^{2}+y^{2}+z^{2}-6 x y+2 y z-6 x z$

(iv) $(m+2 n-5 p)^{2}$

$=m^{2}+(2 n)^{2}+(-5 p)^{2}+2 m \times 2 n$

$+(2 \times 2 n x-5 p)+2 m \times-5 p$

$=m^{2}+4 n^{2}+25 p^{2}+4 m n-20 n p-10 p m$

$(v)(2+x-2 y)^{2}$

$=2^{2}+x^{2}+(-2 y)^{2}+2(2)(x)$

$+2(x)(-2 y)+2(2)(-2 y)$

$=4+x^{2}+4 y^{2}+4 x-4 x y-8 y$

(vi) $\left(a^{2}+b^{2}+c^{2}\right)^{2}$

$=\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+\left(c^{2}\right)^{2}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 a^{2} c^{2}$

$=a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}$

$($ vii $)(a b+b c+c a)^{2}$

$=(a b)^{2}+(b c)^{2}+(c a)^{2}+2(a b)(b c)$

$+2(b c)(c a)+2(a b)(c a)$

$=a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2(a c) b^{2}$

$+2(a b)(c)^{2}+2(b c)(a)^{2}$

(viii) $(x / y+y / z+z / x)^{2}$

$=\left(\frac{x}{y}\right)^{2}+\left(\frac{y}{z}\right)^{2}+\left(\frac{z}{x}\right)^{2}$

$+2 \frac{x}{y} \frac{y}{z}+2 \frac{y}{z} \frac{z}{x}+2 \frac{z}{x} \frac{x}{y}$

$=\left(\frac{x^{2}}{y^{2}}\right)+\left(\frac{y^{2}}{z^{2}}\right)+\left(\frac{z^{2}}{x^{2}}\right)$

$+2 \frac{x}{z}+2 \frac{y}{x}+2 \frac{z}{y}$

$(\mathrm{ix})(\mathrm{a} / \mathrm{bc}+\mathrm{b} / \mathrm{ac}+\mathrm{c} / \mathrm{ab})^{2}$

$=\left(\frac{a}{b c}\right)^{2}+\left(\frac{b}{c a}\right)^{2}+\left(\frac{c}{a b}\right)^{2}+2\left(\frac{a}{b c}\right)\left(\frac{b}{c a}\right)$

$+2\left(\frac{b}{c a}\right)\left(\frac{c}{a b}\right)+2\left(\frac{a}{b c}\right)\left(\frac{c}{a b}\right)$

$=\left(\frac{a^{2}}{b^{2} c^{2}}\right)+\left(\frac{b^{2}}{c^{2} a^{2}}\right)+\left(\frac{c^{2}}{a^{2} b^{2}}\right)$

$+\frac{2}{a^{2}}+\frac{2}{b^{2}}+\frac{2}{c^{2}}$

$(x)(x+2 y+4 z)^{2}$

$=x^{2}+(2 y)^{2}+(4 z)^{2}+(2 x)(2 y)$

$+2(2 y)(4 z)+2 x(4 z)$

$=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z$

$(x i)(2 x-y+z)^{2}$

$=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(2 x)(-y)$

$+2(-y)(z)+2(2 x)(z)$

$=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$

$\left(\right.$ xii) $(-2 x+3 y+2 z)^{2}$

$=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)$

$+2(3 y)(2 z)+2(-2 x)(2 z)$

$=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z$

Question 2: Simplify

(i) $(a+b+c)^{2}+(a-b+c)^{2}$

(ii) $(a+b+c)^{2}-(a-b+c)^{2}$

(iii) $(a+b+c)^{2}+(a-b+c)^{2}+(a+b-c)^{2}$

(iv) $(2 x+p-c)^{2}-(2 x-p+c)^{2}$

$(v)\left(x^{2}+y^{2}-z^{2}\right)^{2}-\left(x^{2}-y^{2}+z^{2}\right)^{2}$

Solution: (i) $(a+b+c)^{2}+(a-b+c)^{2}$

$=\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)$

$+\left(a^{2}+(-b)^{2}+c^{2}-2 a b-2 b c+2 c a\right)$

$=2 a^{2}+2 b^{2}+2 c^{2}+4 c a$

(ii) $(a+b+c)^{2}-(a-b+c)^{2}$

$=\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)$

$-\left(a^{2}+(-b)^{2}+c^{2}-2 a b-2 b c+2 c a\right)$

$=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$

$-a^{2}-b^{2}-c^{2}+2 a b+2 b c-2 c a$

$=4 a b+4 b c$

(iii) $(a+b+c)^{2}+(a-b+c)^{2}+(a+b-c)^{2}$

$=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$

$+\left(a^{2}+b^{2}+(c)^{2}-2 a b-2 c b+2 c a\right)$

$+\left(a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a\right)$

$=3 a^{2}+3 b^{2}+3 c^{2}+2 a b-2 b c+2 c a$

(iv) $(2 x+p-c)^{2}-(2 x-p+c)^{2}$

$=\left[4 x^{2}+p^{2}+c^{2}+4 x p-2 p c-4 x c\right]$

$-\left[4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 x c\right]$

$=4 x^{2}+p^{2}+c^{2}+4 x p-2 p c-4 c x$

$-4 x^{2}-p^{2}-c^{2}+4 x p+2 p c-4 c x$

$=8 x p-8 x$

$=8(\mathrm{xp}-\mathrm{xc})$

(v) $\left(\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{z}^{2}\right)^{2}-\left(\mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{Z}^{2}\right)^{2}$

$=\left(x^{2}+y^{2}+(-z)^{2}\right)^{2}-\left(x^{2}-y^{2}+z^{2}\right)^{2}$

$=\left[x^{4}+y^{4}+z^{4}+2 x^{2} y^{2}-2 y^{2} z^{2}-2 x^{2} Z^{2}\right.$

$-\left[x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}+2 x^{2} z^{2}\right]$

$=4 x^{2} y^{2}-4 z^{2} x^{2}$

Question 3: If $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=16$, find the value of $a b+b c+c a$

Solution. $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=16$ (given)

Choose $a+b+c=0$

Squaring both sides,

$(a+b+c)^{2}=0$

$a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=0$

$16+2(a b+b c+c)=0$

$2(a b+b c+c a)=-16$

$a b+b c+c a=-16 / 2=-8$

or $a b+b c+c a=-8$

Exercise 4.3 Page No: $4.19$

Question 1: Find the cube of each of the following binomial expressions:

(i) $(1 / x+y / 3)$

(ii) $\left(3 / x-2 / x^{2}\right)$

(iii) $(2 x+3 / x)$

(iv) $(4-1 / 3 x)$

Solution: $\left[U\right.$ sing identities: $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ and $\left.(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\right]$

(i)$\left.\left(\frac{1}{x}+\frac{y}{3}\right)\right)^{3}=\left(\frac{1}{x}\right)^{3}+\left(\frac{y}{3}\right)^{3}$

$+3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)\left(\frac{1}{x}+\frac{y}{3}\right)$

$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+3 \times \frac{1}{x} \times \frac{y}{3}\left(\frac{1}{x}+\frac{y}{3}\right)$

$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\left(\frac{y}{x} \times \frac{1}{x}\right)+\left(\frac{y}{x} \times \frac{y}{3}\right)$

$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}$

(ii) $\left(\frac{3}{x}-\frac{2}{x^{2}}\right)^{3}=\left(\frac{3}{x}\right)^{3}-\left(\frac{2}{x^{2}}\right)^{3}$

$-3\left(\frac{3}{x}\right)\left(\frac{2}{x^{2}}\right)\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$

$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-3 \times \frac{3}{x}$

x $\frac{2}{x^{2}}\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$

$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{18}{x^{3}}\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$

$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{54}{x^{4}}+\frac{36}{x^{5}}$

(iii) $\left(2 x+\frac{3}{x}\right)^{3}$

$=8 x^{3}+\frac{27}{x^{3}}+\frac{18 x}{x}\left(2 x+\frac{3}{x}\right)$

$=8 x^{3}+\frac{27}{x^{3}}+\frac{18 x}{x}\left(2 x+\frac{3}{x}\right)$

$=8 x^{3}+\frac{27}{x^{3}}$

$+(18 \times 2 x)+\left(18 \times \frac{3}{x}\right)$

$=8 x^{3}+\frac{27}{x^{3}}+36 x+\frac{54}{x}$

(iv) $\left(4-\frac{1}{3 x}\right)^{3}=4^{3}-\left(\frac{1}{3 x}\right)^{3}$

$-3(4)\left(\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$

$=64-\frac{1}{27 x^{3}}-\frac{4}{x}\left(4-\frac{1}{3 x}\right)$

$=64-\frac{1}{27 x^{3}}-\frac{16}{x}+\frac{4}{3 x^{2}}$

Question 2: Simplify each of the following:

(i) $(x+3)^{3}+(x-3)^{3}$

(ii) $(x / 2+y / 3)^{3}-(x / 2-y / 3)^{3}$

(iii) $(x+2 / x)^{3}+(x-2 / x)^{3}$

(iv) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$

Solution: [Using identities:

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

$(a+b)(a-b)=a^{2}-b^{2}$

$(a+b)^{2}=a^{2}+b^{2}+2 a b$ and

$\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

(i) $(x+3)^{3}+(x-3)^{3}$

Here $a=(x+3), b=(x-3)$

$=(x+3+x-3)\left[(x+3)^{2}\right.$

$\left.+(x-3)^{2}-(x+3)(x-3)\right]$

$=2 x\left[\left(x^{2}+9+6 x\right)\right.$

$\left.+\left(x^{2}+9-6 x\right)-x^{2}+9\right]$

$=2 x\left[\left(x^{2}+9+6 x+x^{2}\right.\right.$

$\left.\left.+9-6 x-x^{2}+9\right)\right]$

$=2 x\left(x^{2}+27\right)$

$=2 x^{3}+54 x$

(ii) $(x / 2+y / 3)^{3}-(x / 2-y / 3)^{3}$

Here $a=(x / 2+y / 3)$ and $b=(x / 2-y / 3)$

$=\left[\left(\frac{x}{2}+\frac{y}{3}\right)-\left(\frac{x}{2}-\frac{y}{3}\right)\right]$

$\left[\left(\frac{x}{2}+\frac{y}{3}\right)^{2}+\left(\frac{x}{2}-\frac{y}{3}\right)^{2}+\left(\frac{x}{2}+\frac{y}{3}\right)\left(\frac{x}{2}-\frac{y}{3}\right)\right]$

$=\frac{2 y}{3}\left[\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{2 x y}{6}\right)\right.$

$\left.+\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}-\frac{2 x y}{6}\right)+\frac{x^{2}}{4}-\frac{y^{2}}{9}\right]$

$=\frac{2 y}{3}\left[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{x^{2}}{4}+\frac{x^{2}}{4}\right]$

$=\frac{2 y}{3}\left[\frac{3 x^{2}}{4}+\frac{y^{2}}{9}\right]$

$=\frac{x^{2} y}{2}+\frac{2 y^{3}}{27}$

(iii) $(x+2 / x)^{3}+(x-2 / x)^{3}$

Here $a=(x+2 / x)$ and $b=(x-2 / x)$

$=\left(x+\frac{2}{x}+x-\frac{2}{x}\right)$

$\left[\left(x+\frac{2}{x}\right)^{2}+\left(x-\frac{2}{x}\right)^{2}-\left(\left(x+\frac{2}{x}\right)\left(x-\frac{2}{x}\right)\right)\right]$

$=(2 x)\left[\left(x^{2}+\frac{4}{x^{2}}+\frac{4 x}{x}\right)\right.$

$+\left(x^{2}+\frac{4}{x^{2}}-\frac{4 x}{x}\right)-\left(x^{2}-\frac{4}{x^{2}}\right)$

$=(2 x)\left[\left(x^{2}+\frac{4}{x^{2}}+\frac{4}{x^{2}}+\frac{4}{x^{2}}\right)\right.$

$=(2 x)\left[\left(x^{2}+\frac{12}{x^{2}}\right)\right.$

$=2 x^{3}+\frac{24}{x}$

(iv) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$

Here $a=(2 x-5 y)$ and $b=2 x+5 y$

$=(2 x-5 y-2 x-5 y)\left[(2 x-5 y)^{2}\right.$

$\left.+(2 x+5 y)^{2}+((2 x-5 y)(2 x+5 y))\right]$

$=(-10 y)\left[\left(4 x^{2}+25 y^{2}-20 x y\right)\right.$

$\left.+\left(4 x^{2}+25 y^{2}+20 x y\right)+4 x^{2}-25 y^{2}\right]$

$=(-10 y)\left[4 x^{2}+4 x^{2}+4 x^{2}+25 y^{2}\right]$

$=(-10 y)\left[12 x^{2}+25 y^{2}\right\}$

$=-120 x^{2} y-250 y^{3}$

Question 3 : If $a+b=10$ and $a b=21$, find the value of $a^{3}+b^{3}$.

Solution: $a+b=10, a b=21 \text { (given) }$

Choose $a+b=10$

Cubing both sides,

$(a+b)^{3}=(10)^{3}$

$a^{3}+b^{3}+3 a b(a+b)=1000$

$a^{3}+b^{3}+3 \times 21 \times 10$

$=1000$ (using given values)

$a^{3}+b^{3}+630=1000$

$a^{3}+b^{3}=1000-630=370$

or $a^{3}+b^{3}=370$

Question 4: If $a-b=4$ and $a b=21$, find the value of $a^{3}-b^{3}$

Solution: $a-b=4, a b=21$ (given)

Choose $a-b=4$

Cubing both sides,

$(a-b)^{3}=(4)^{3}$

$a^{3}-b^{3}-3 a b(a-b)=64$

$a^{3}-b^{3}-3 \times 21 \times 4=64$ (using given values)

$a^{3}-b^{3}-252=64$

$a^{3}-b^{3}=64+252$

$=316$

Or $a^{3}-b^{3}=316$

Question 5: If $x+1 / x=5$, find the value of $x^{3}+1 / x^{3}$.

Solution: Given: $x+1 / x=5$

Apply Cube on $x+1 / x$

$\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}$

$+3\left(x \times \frac{1}{x}\right)\left(x+\frac{1}{x}\right)$

$5^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)$

$125=x^{3}+\frac{1}{x^{3}}+3(5)$

$125=x^{3}+\frac{1}{x^{3}}+15$

$125-15=x^{3}+\frac{1}{x^{3}}$

$x^{3}+\frac{1}{x^{3}}=110$

Question 6: If $x-1 / x=7$, find the value of $x^{3}-1 / x^{3}$.

Solution: Given: $x-1 / x=7$

Apply Cube on $x-1 / x$

$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}$

$-3\left(x \times \frac{1}{x}\right)\left(x-\frac{1}{x}\right)$

$7^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$ $343=x^{3}-\frac{1}{x^{3}}-(3 \times 7)$

$343+21=x^{3}-\frac{1}{x^{3}}$

$x^{3}-\frac{1}{x^{3}}=364$

Question 7: If $x-1 / x=5$, find the value of $x^{3}-1 / x^{3}$

Solution: Given: $x-1 / x=5$

Apply Cube on $\mathrm{x}-1 / \mathrm{x}$

$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}$

$-3\left(x \times \frac{1}{x}\right)\left(x-\frac{1}{x}\right)$

$5^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$

$125=x^{3}-\frac{1}{x^{3}}-(3 \times 5)$

$125=x^{3}-\frac{1}{x^{3}}-15$

$125+15=x^{3}-\frac{1}{x^{3}}$

$x^{3}-\frac{1}{x^{3}}=140$

Question 8: If $\left(x^{2}+1 / x^{2}\right)=51$, find the value of $x^{3}-1 / x^{3}$

Solution: We know that: $(x-y)^{2}=x^{2}+y^{2}-2 x y$

Replace $y$ with $1 / x$, we get

$(x-1 / x)^{2}=x^{2}+1 / x^{2}-2$

Since $\left(x^{2}+1 / x^{2}\right)=51$ (given)

$(x-1 / x)^{2}=51-2=49$

or $(x-1 / x)=\pm 7$

Now, Find $x^{3}-1 / x^{3}$

We know that, $x^{3}-y^{3}=(x-y)\left(x^{2}+y^{2}+x y\right)$

Replace $\mathrm{y}$ with $1 / \mathrm{x}$, we get

$x^{3}-1 / x^{3}=(x-1 / x)\left(x^{2}+1 / x^{2}+1\right)$

Use $(x-1 / x)=7$ and $\left(x^{2}+1 / x^{2}\right)=51$

$x^{3}-1 / x^{3}=7 \times 52=364$

$x^{3}-1 / x^{3}=364$

Question 9: If $\left(x^{2}+1 / x^{2}\right)=98$, find the value of $x^{3}+1 / x^{3}$

Solution: We know that: $(x+y)^{2}=x^{2}+y^{2}+2 x y$

Replace $y$ with $1 / x$, we get

$ (x+1 / x)^{2}=x^{2}+1 / x^{2}+2$

Since $\left(x^{2}+1 / x^{2}\right)=98$ (given)

$(x+1 / x)^{2}=98+2=100$

or $(x+1 / x)=\pm 10$

Now, Find $x^{3}+1 / x^{3}$

We know that, $x^{3}+y^{3}=(x+y)\left(x^{2}+y^{2}-x y\right)$

Replace $y$ with $1 / x$, we get

$x^{3}+1 / x^{3}=(x+1 / x)\left(x^{2}+1 / x^{2}-1\right)$

Use $(x+1 / x)=10$ and $\left(x^{2}+1 / x^{2}\right)=98$

$x^{3}+1 / x^{3}=10 \times 97=970$

$\mathrm{X}^{3}+1 / \mathrm{X}^{3}=970$

Question 10: If $2 x+3 y=13$ and $x y=6$, find the value of $8 x^{3}+27 y^{3}$

Solution: Given: $2 x+3 y=13, x y=6$

Cubing $2 x+3 y=13$ both sides, we get

$(2 x+3 y)^{3}=(13)^{3}$

$(2 x)^{3}+(3 y)^{3}+3(2 x)(3 y)(2 x+3 y)=2197$

$8 x^{3}+27 y^{3}+18 x y(2 x+3 y)=2197$

$8 x^{3}+27 y^{3}+18 \times 6 \times 13=2197$

$8 x^{3}+27 y^{3}+1404=2197$

$8 x^{3}+27 y^{3}=2197-1404=793$

$8 x^{3}+27 y^{3}=793$

Question 11: If $3 x-2 y=11$ and $x y=12$, find the value of $27 x^{3}-8 y^{3}$

Solution. Given: $3 x-2 y=11$ and $x y=12$

Cubing $3 x-2 y=11$ both sides, we get

$(3 x-2 y)^{3}=(11)^{3}$

$(3 x)^{3}-(2 y)^{3}-3(3 x)(2 y)(3 x-2 y)=1331$

$27 x^{3}-8 y^{3}-18 x y(3 x-2 y)=1331$

$27 x^{3}-8 y^{3}-18 \times 12 \times 11=1331$

$27 x^{3}-8 y^{3}-2376=1331$

$27 x^{3}-8 y^{3}=1331+2376=3707$

$27 x^{3}-8 y^{3}=3707$

Exercise $4.4$ Page No: $4.23$

Question 1: Find the following products:

(i) $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$

(ii) $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

(iii) $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$

(iv) $(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$

(v) $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$

(vi) $(3+5 / x)\left(9-15 / x+25 / x^{2}\right)$

$\left(\right.$ vii) $(2 / x+3 x)\left(4 / x^{2}+9 x^{2}-6\right)$

(viii) $\left(3 / x-2 x^{2}\right)\left(9 / x^{2}+4 x^{4}-6 x\right)$

$($ ix $)(1-x)\left(1+x+x^{2}\right)$

$(x)(1+x)\left(1-x+x^{2}\right)$

$(x i)\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$

$(x i i)\left(x^{3}+1\right)\left(x^{6}-x^{3}+1\right)$

Solution: (i) $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$

$\left.=(3 x+2 y)\left[(3 x)^{2}-(3 x)(2 y)+(2 y)^{2}\right)\right]$

We know, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

$=(3 x)^{3}+(2 y)^{3}$

$=27 x^{3}+8 y^{3}$

(ii) $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

$\left.=(4 x-5 y)\left[(4 x)^{2}+(4 x)(5 y)+(5 y)^{2}\right)\right]$

We know, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

$=(4 x)^{3}-(5 y)^{3}$

$=64 x^{3}-125 y^{3}$

(iii) $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$

$\left.=\left(7 p^{4}+q\right)\left[\left(7 p^{4}\right)^{2}-\left(7 p^{4}\right)(q)+(q)^{2}\right)\right]$

We know, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

$=\left(7 p^{4}\right)^{3}+(q)^{3}$

$=343 \mathrm{p}^{12}+\mathrm{q}^{3}$

(iv) $(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$

We know, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

$(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$

$=\left(\frac{x}{2}+2 y\right)\left[\left(\frac{x}{2}\right)^{2}-\frac{x}{2}(2 y)+(2 y)^{2}\right]$

$=\left(\frac{x}{2}\right)^{3}+(2 y)^{3}$

$=\frac{x^{3}}{8}+8 y^{3}$

(v) $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$

$=\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{3}{x}\right)^{2}+\left(\frac{5}{y}\right)^{2}+\left(\frac{3}{x}\right)\left(\frac{5}{y}\right)$

$=\left(\frac{3}{x}\right)^{3}-\left(\frac{5}{y}\right)^{3}$

$=\left(\frac{27}{x^{3}}\right)-\left(\frac{125}{y^{3}}\right)$

$\left[\right.$ Using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

(vi) $(3+5 / x)\left(9-15 / x+25 / x^{2}\right)$

$=\left(3+\frac{5}{x}\right)\left[\left(3^{2}\right)-3\left(\frac{5}{x}\right)+\left(\frac{5}{x}\right)^{2}\right]$

$=(3)^{3}+\left(\frac{5}{x}\right)^{3}$

$=27+\frac{125}{x^{3}}$

$\left[\right.$ Using: $\left.a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

(vii) $(2 / x+3 x)\left(4 / x^{2}+9 x^{2}-6\right)$

$=\left(\frac{2}{x}+3 x\right)\left[\left(\frac{2}{x}\right)^{2}\right.$ $\left.+(3 x)^{2}-\left(\frac{2}{x}\right)(3 x)\right]$

$=\left(\frac{2}{x}\right)^{3}+(3 x)^{3}$

$=\frac{8}{x^{3}}+27 x^{3}$

$\left[\right.$ Using: $\left.a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

(viii) $\left(3 / x-2 x^{2}\right)\left(9 / x^{2}+4 x^{4}-6 x\right)$

$=\left(\frac{3}{x}-2 x^{2}\right)\left[\left(\frac{3}{x}\right)^{2}\right.$

$\left.+\left(2 x^{2}\right)^{2}-\left(\frac{3}{x}\right)\left(2 x^{2}\right)\right]$

$=\left(\frac{3}{x}-2 x^{2}\right)\left[\left(\frac{9}{x^{2}}\right)\right.$

$\left.+4 x^{4}-\left(\frac{3}{x}\right)\left(2 x^{2}\right)\right]$

$=\left(\frac{3}{x}\right)^{3}-\left(2 x^{2}\right)^{3}$

$=\frac{27}{x^{3}}-8 x^{6}$

$\left[\right.$ Using $\left.: a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

$(i x)(1-x)\left(1+x+x^{2}\right)$

And we know, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

$(1-x)\left(1+x+x^{2}\right)$ can be written as

$(1-x)\left[\left(1^{2}+(1)(x)+x^{2}\right)\right]$

$=(1)^{3}-(x)^{3}$

$=1-x^{3}$

$(x)(1+x)\left(1-x+x^{2}\right)$

And we know, $\left.a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

$(1+x)\left(1-x+x^{2}\right)$ can be written as

$(1+x)\left[\left(1^{2}-(1)(x)+x^{2}\right)\right]$

$=(1)^{3}+(x)^{3}$

$=1+x^{3}$

(xi) $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$ can be written as

$\left(x^{2}-1\right)\left[\left(x^{2}\right)^{2}-1^{2}+\left(x^{2}\right)(1)\right]$

$=\left(x^{2}\right)^{3}-1^{3}$

$=x^{6}-1$

$\left[\right.$ using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

(xii) $\left(x^{3}+1\right)\left(x^{6}-x^{3}+1\right)$ can be written as,

$\left(x^{3}+1\right)\left[\left(x^{3}\right)^{2}-\left(x^{3}\right)(1)+1^{2}\right]$

$=\left(\mathrm{X}^{3}\right)^{3}+1^{3}$

$=x^{9}+1$

[using $\left.a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

Question 2: If $x=3$ and $y=-1$, find the values of each of the following using in identity:

(i) $\left(9 y^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$

(ii) $(3 / x-x / 3)\left(x^{2} / 9+9 / x^{2}+1\right)$

(iii) $(x / 7+y / 3)\left(x^{2} / 49+y^{2} / 9-x y / 21\right)$

(iv) $(x / 4-y / 3)\left(x^{2} / 16+x y / 12+y^{2} / 9\right)$

(v) $(5 / x+5 x)\left(25 / x^{2}-25+25 x^{2}\right)$

Solution: (i) $\left(9 y^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$

$=\left(9 y^{2}-4 x^{2}\right)\left[\left(9 y^{2}\right)^{2}+9 y^{2} \times 4 x^{2}+\left(4 x^{2}\right)^{2}\right]$

$=\left(9 y^{2}\right)^{3}-\left(4 x^{2}\right)^{3}$

$=729 y^{6}-64 x^{6}$

Put $x=3$ and $y=-1$

$=729-46656$

$=-45927$

(ii) Put $x=3$ and $y=-1$

$(3 / x-x / 3)\left(x^{2} / 9+9 / x^{2}+1\right)$

$=\left(\frac{3}{x}-\frac{x}{3}\right)\left[\left(\frac{x}{3}\right)^{2}+\frac{x}{3} \times \frac{3}{x}+\left(\frac{3}{x}\right)^{2}\right]$

$=\left(\frac{3}{x}\right)^{3}-\left(\frac{x}{3}\right)^{3}$

$=\left(\frac{3}{3}\right)^{3}-\left(\frac{3}{3}\right)^{3}$

$=1^{3}-1^{3}=0$

(iii) Put $x=3$ and $y=-1$

$(x / 7+y / 3)\left(x^{2} / 49+y^{2} / 9-x y / 21\right)$

$=\left(\frac{x}{7}+\frac{y}{3}\right)\left[\left(\frac{x}{7}\right)^{2}-\frac{x}{7} \times \frac{y}{3}-\left(\frac{y}{3}\right)^{2}\right]$

$=\left(\frac{x}{7}\right)^{3}+\left(\frac{y}{3}\right)^{3}$

$=\frac{x^{3}}{343}+\frac{y^{3}}{27}$

$=\frac{(3)^{3}}{343}+\frac{(-1)^{3}}{27}$

$=\frac{27}{343}-\frac{1}{27}$

$=\frac{729-343}{9261}:=\frac{386}{9261}$

(iv) Put $x=3$ and $y=-1$

$(x / 4-y / 3)\left(x^{2} / 16+x y / 12+y^{2} / 9\right)$

$=\left(\frac{x}{4}-\frac{y}{3}\right)\left[\left(\frac{x}{4}\right)^{2}+\frac{x}{4} \times \frac{y}{3}+\left(\frac{y}{3}\right)^{2}\right]$

$=\left(\frac{x}{4}\right)^{3}-\left(\frac{y}{3}\right)^{3}$

$=\frac{x^{3}}{64}-\frac{y^{3}}{27}$

$=\frac{(3)^{3}}{64}-\frac{(-1)^{3}}{27}$

$=\frac{27}{64}+\frac{1}{27}$

$=\frac{793}{1728}$

(v) Put $x=3$ and $y=-1$

$(5 / x+5 x)\left(25 / x^{2}-25+25 x^{2}\right)$

$=\left(\frac{5}{x}+5 x\right)\left[\left(\frac{5}{x}\right)^{2}-\frac{5}{x} \times 5 x+(5 x)^{2}\right]$

$=\left(\frac{5}{x}\right)^{3}+(5 x)^{3}$

$=\frac{125}{x^{3}}+125 x^{3}$

$=\frac{125}{(3)^{3}}+125 \times(3)^{3}$

$=\frac{125}{27}+125 \times 27$

$=\frac{125}{27}+3375$

$=\frac{91250}{27}$

Question 3: If $a+b=10$ and $a b=16$, find the value of $a^{2}-a b+b^{2}$ and $a^{2}+a b+b^{2}$

Solution: $a+b=10, a b=16$

Squaring, $a+b=10$, both sides

$(a+b)^{2}=(10)^{2}$

$a^{2}+b^{2}+2 a b=100$

$a^{2}+b^{2}+2 \times 16=100$

$a^{2}+b^{2}+32=100$

$a^{2}+b^{2}=100-32=68$

$a^{2}+b^{2}=68$

Again, $a^{2}-a b+b^{2}=a^{2}+b^{2}-a b=68-16=52$ and

$a^{2}+a b+b^{2}=a^{2}+b^{2}+a b=68+16=84$

Question 4 : If $a+b=8$ and $a b=6$, find the value of $a^{3}+b^{3}$

Solution: $a+b=8, a b=6$

Cubing, $a+b=8$, both sides, we get

$(a+b)^{3}=(8)^{3}$

$a^{3}+b^{3}+3 a b(a+b)=512$

$a^{3}+b^{3}+3 \times 6 \times 8=512$ $a^{3}+b^{3}+144=512$

$a^{3}+b^{3}=512-144=368$

$a^{3}+b^{3}=368$

Exercise $4.5$ Page No: $4.28$

Question 1: Find the following products:

(i) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$

(ii) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$

(iii) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$

(iv) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$

Solution: (i) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$

$=(3 x+2 y+2 z)\left[(3 x)^{2}+(2 y)^{2}+(2 z)^{2}\right.$

$-3 x \times 2 y-2 y \times 2 z-2 z \times 3 x]$ $=(3 x)^{3}+(2 y)^{3}+(2 z)^{3}-3 \times 3 x \times 2 y \times 2 z$

$=27 x^{3}+8 y^{3}+8 Z^{3}-36 x y z$

(ii) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$

$=(4 x-3 y+2 z)\left[(4 x)^{2}+(-3 y)^{2}+(2 z)^{2}\right.$

$-4 x \times(-3 y)-(-3 y) \times(2 z)-(2 z \times 4 x)]$

$=(4 x)^{3}+(-3 y)^{3}+(2 z)^{3}-3 \times 4 x \times(-3 y) x(2 z)$

$=64 x^{3}-27 y^{3}+8 z^{3}+72 x y z$

(iii) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$

$=(2 a-3 b-2 c)\left[(2 a)^{2}+(-3 b)^{2}+(-2 c)^{2}\right.$

$-2 a \times(-3 b)-(-3 b) \times(-2 c)-(-2 c) \times 2 a]$

$=(2 a)^{3}+(-3 b)^{3}+(-2 c)^{3}-3 \times 2 a \times(-3 b)(-2 c)$

$=8 a^{3}-21 b^{3}-8 c^{3}-36 a b c$

(iv) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$

$=[3 x+(-4 y)+5 z]\left[(3 x)^{2}+(-4 y)^{2}+(5 z)^{2}\right.$

$-3 x \times(-4 y)-(-4 y)(5 z)-5 z \times 3 x]$

$=(3 x)^{3}+(-4 y)^{3}+(5 z)^{3}-3 \times 3 x \times(-4 y)$

$=27 x^{3}-64 y^{3}+125 z^{3}+180 x y z$

Question 2: If $x+y+z=8$ and $x y+y z+z x=20$, find the value of $x^{3}+y^{3}+z^{3}-3 x y z$

Solution: We know, $x^{3}+y^{3}+z^{3}-3 x y z$

$=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

Squaring, $x+y+z=8$ both sides, we get

$(x+y+z)^{2}=(8)^{2}$

$x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=64$

$x^{2}+y^{2}+z^{2}+2 \times 20=64$

$x^{2}+y^{2}+z^{2}+40=64$

$x^{2}+y^{2}+z^{2}=24$

Now,

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)$

$\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right]$

$=8(24-20)$

$=8 \times 4$

$=32$ $\Rightarrow x^{3}+y^{3}+z^{3}-3 x y z=32$

Question 3: If $a+b+c=9$ and $a b+b c+c a=26$, find the value of $a^{3}+b^{3}+c^{3}-3 a b c$

Solution: $a+b+c=9, a b+b c+c a=26$

Squaring, $a+b+c=9$ both sides, we get

$(a+b+c)^{2}=(9)^{2}$

$a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=81$

$a^{2}+b^{2}+c^{2}+2 \times 26=81$

$a^{2}+b^{2}+c^{2}+52=81$

$a^{2}+b^{2}+c^{2}=29$

Now, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right]\right.$

$=9[29-26]$

$=9 \times 3$

$=27$

$\Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=27$

Exercise VSAQs Page No: $4.28$

Question 1 : If $x+1 / x=3$, then find the value of $x^{2}+1 / x^{2}$

Solution: $x+1 / x=3$

Squaring both sides, we have

$(x+1 / x)^{2}=3^{2}$

$x^{2}+1 / x^{2}+2=9$

$x^{2}+1 / x^{2}=9-2=7$

Question 2: If $x+1 / x=3$, then find the value of $x^{\wedge} 6+1 / x^{\wedge} 6$

Solution: $x+1 / x=3$

Squaring both sides, we have

$(x+1 / x)^{2}=3^{2}$

$x^{2}+1 / x^{2}+2=9$

$x^{2}+1 / x^{2}=9-2=7$

$x^{2}+1 / x^{2}=7 \ldots$(1)

Cubing equation (1) both sides,

$=\left(x^{2}+\frac{1}{x^{2}}\right)^{3}=(7)^{3}$

$=x^{6}+\frac{1}{x^{6}}+3\left(x^{2}+\frac{1}{x^{2}}\right)=343$

$=x^{6}+\frac{1}{x^{6}}+3 \times 7=343$

$=x^{6}+\frac{1}{x^{6}}=322$

Question 3: If $a+b=7$ and $a b=12$, find the value of $a^{2}+b^{2}$

Solution: $a+b=7, a b=12$

Squaring, $a+b=7$, both sides,

$(a+b)^{2}=(7)^{2}$

$a^{2}+b^{2}+2 a b=49$

$a^{2}+b^{2}+2 \times 12=49$

$a^{2}+b^{2}+24=49$

$a^{2}+b^{2}=25$

Question 4: If $a-b=5$ and $a b=12$, find the value of $a^{2}+b^{2}$

Solution: $a-b=5, a b=12$

Squaring, $a-b=5$, both sides,

$(a-b)^{2}=(5)^{2}$

$a^{2}+b^{2}-2 a b=25$

$a^{2}+b^{2}-2 \times 12=25$

$a^{2}+b^{2}-24=25$

$a^{2}+b^{2}=49$

Also Read,

Download NCERT Class 9 Maths Chapterwise Book Free

Download NCERT Class 10 Maths Chapterwise Book Free

Download NCERT Class 9 Maths Chapterwise Exemplar Free

Download NCERT Class 10 Maths Chapterwise Exemplar Free

If you have any Confusion related to RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 9 by Kota’s top Faculties Install the eSaral App