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

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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,

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

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

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

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

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

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