Exercise 2.4 (Revised) - Chapter 2 - Polynomials - Ncert Solutions class 9 - Maths
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Chapter 2 - Polynomials | NCERT Solutions for Class 9 Maths
Ex 2.4 Question 1.
Use suitable identities to find the following products:
(i) $(x+4)(x+10)$
(ii) $(x+8)(x-10)$
(iii) $(3 x+4)(3 x-5)$
(iv) $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$
(v) $(3-2 x)(3+2 x)$
Answer.
(i) $(x+4)(x+10)$
We know that $(x+a)(x+b)=x^2+(a+b) x+a b$.
We need to apply the above identity to find the product $(x+4)(x+10)$
$
\begin{aligned}
& (x+4)(x+10)=x^2+(4+10) x+(4 \times 10) \\
& =x^2+14 x+40
\end{aligned}
$
Therefore, we conclude that the product $(x+4)(x+10)$ is $x^2+14 x+40$.
(ii) $(x+8)(x-10)$
We know that $(x+a)(x+b)=x^2+(a+b) x+a b$.
We need to apply the above identity to find the product $(x+8)(x-10)$
$
\begin{aligned}
& (x+8)(x-10)=x^2+[8+(-10)] x+[8 \times(-10)] \\
& =x^2-2 x-80 .
\end{aligned}
$
Therefore, we conclude that the product $(x+8)(x-10)$ is $x^2-2 x-80$.
(iii) $(3 x+4)(3 x-5)$
We know that $(x+a)(x+b)=x^2+(a+b) x+a b$.
We need to apply the above identity to find the product $(3 x+4)(3 x-5)$
$
\begin{aligned}
& (3 x+4)(3 x-5)=(3 x)^2+[4+(-5)] 3 x+[4 \times(-5)] \\
& =9 x^2-3 x-20 .
\end{aligned}
$
Therefore, we conclude that the product $(3 x+4)(3 x-5)$ is $9 x^2-3 x-20$.
(iv) $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$
We know that $(x+y)(x-y)=x^2-y^2$.
We need to apply the above identity to find the product $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$
$
\begin{aligned}
& \left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right) \\
& =\left(y^2\right)^2-\left(\frac{3}{2}\right)^2=y^4-\frac{9}{4} .
\end{aligned}
$
Therefore, we conclude that the product $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$ is $\left(y^4-\frac{9}{4}\right)$.
(v) $(3+2 x)(3-2 x)$
We know that $(x+y)(x-y)=x^2-y^2$.
We need to apply the above identity to find the product $(3+2 x)(3-2 x)$
$
\begin{aligned}
& (3+2 x)(3-2 x)=(3)^2-(2 x)^2 \\
& =9-4 x^2 .
\end{aligned}
$
$\text { Therefore, we conclude that the product }(3+2 x)(3-2 x) \text { is }\left(9-4 x^2\right) \text {. }$
Ex 2.4 Question 2.
Evaluate the following products without multiplying directly:
(i) $103 \times 107$
(ii) $98 \times 96$
(iii) $104 \times 96$
Answer.
(i) $103 \times 107$
$103 \times 107$ can al so be written as $(100+3)(100+7)$.
We can observe that, we can apply the identity $(x+a)(x+b)=x^2+(a+b) x+a b$
$
\begin{aligned}
& (100+3)(100+7)=(100)^2+(3+7)(100)+3 \times 7 \\
& =10000+1000+21 \\
& =11021
\end{aligned}
$
Therefore, we conclude that the value of the product $103 \times 107$ is 11021 .
(ii) $95 \times 96$
$95 \times 96$ can also be written as $(100-5)(100-4)$
We can observe that, we can apply the identity $(x+a)(x+b)=x^2+(a+b) x+a b$
$
\begin{aligned}
& (100-5)(100-4)=(100)^2+[(-5)+(-4)](100)+(-5) \times(-4) \\
& =10000-900+20 \\
& =9120
\end{aligned}
$
Therefore, we conclude that the value of the product $95 \times 96$ is 9120 .
(iii) $104 \times 96$
$104 \times 96$ can also be written as $(100+4)(100-4)$.
We can observe that, we can apply the identity $(x+y)(x-y)=x^2-y^2$ with respect to the expression $(100+4)(100-4)$, to get
$
\begin{aligned}
& (100+4)(100-4)=(100)^2-(4)^2 \\
& =10000-16 \\
& =9984
\end{aligned}
$
Therefore, we conclude that the value of the product $104 \times 96$ is 9984 .
Ex 2.4 Question 3.
Factorize the following using appropriate identities:
(i) $9 x^2+6 x y+y^2$
(ii) $4 y^2-4 y+1$
(iii) $x^2-\frac{y^2}{100}$
Answer.
(i) $9 x^2+6 x y+y^2$
$
9 x^2+6 x y+y^2=(3 x)^2+2 \times 3 x \times y+(y)^2
$
We can observe that, we can apply the identity $(x+y)^2=x^2+2 x y+y^2$
$
\Rightarrow(3 x)^2+2 \times 3 x \times y+(y)^2=(3 x+y)^2 \text {. }
$
(ii) $4 y^2-4 y+1$
$
4 y^2-4 y+1=(2 y)^2-2 \times 2 y \times 1+(1)^2
$
We can observe that, we can apply the identity $(x-y)^2=x^2-2 x y+y^2$
$
\Rightarrow(2 y)^2-2 \times 2 y \times 1+(1)^2=(2 y-1)^2 \text {. }
$
(iii) $x^2-\frac{y^2}{100}$
We can observe that, we can apply the identity $(x)^2-(y)^2=(x+y)(x-y)$
$
\Rightarrow(x)^2-\left(\frac{y}{10}\right)^2=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)
$
Ex 2.4 Question 4.
Expand each of the following, using suitable identities:
(i) $(x+2 y+4 z)^2$
(ii) $(2 x-y+z)^2$
(iii) $(-2 x+3 y+2 z)^2$
(iv) $(3 a-7 b-c)^2$
(v) $(-2 x+5 y-3 z)^2$
(vi) $\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^2$
Answer.
(i) $(x+2 y+4 z)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
We need to apply the above identity to expand the expression $(x+2 y+4 z)^2$.
$
\begin{aligned}
(x+2 y+4 z)^2 & =(x)^2+(2 y)^2+(4 z)^2+2 \times x \times 2 y+2 \times 2 y \times 4 z+2 \times 4 z \times x \\
& =x^2+4 y^2+16 z^2+4 x y+16 y z+8 z x
\end{aligned}
$
(ii) $(2 x-y+z)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
We need to apply the above identity to expand the expression $(2 x-y+z)^2$.
$
\begin{aligned}
(2 x-y+z)^2 & =[2 x+(-y)+z]^2 \\
& =(2 x)^2+(-y)^2+(z)^2+2 \times 2 x \times(-y)+2 \times(-y) \times z+2 \times z \times 2 x \\
& =4 x^2+y^2+z^2-4 x y-2 y z+4 z x
\end{aligned}
$
(iii) $(-2 x+3 y+2 z)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
We need to apply the above identity to expand the expression $(-2 x+3 y+2 z)^2$.
$
\begin{aligned}
(-2 x+3 y+2 z)^2 & =[(-2 x)+3 y+2 z]^2 \\
& =(-2 x)^2+(3 y)^2+(2 z)^2+2 \times(-2 x) \times 3 y+2 \times 3 y \times 2 z+2 \times 2 z \times(-2 x) \\
& =4 x^2+9 y^2+4 z^2-12 x y+12 y z-8 z x
\end{aligned}
$
(iv) $(3 a-7 b-c)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
We need to apply the above identity to expand the expression $(3 a-7 b-c)^2$.
$
\begin{aligned}
(3 a-7 b-c)^2 & =[3 a+(-7 b)+(-c)]^2 \\
& =(3 a)^2+(-7 b)^2+(-c)^2+2 \times 3 a \times(-7 b)+2 \times(-7 b) \times(-c)+2 \times(-c) \times 3 a \\
& =9 a^2+49 b^2+c^2-42 a b+14 b c-6 a c
\end{aligned}
$
(v) $(-2 x+5 y-3 z)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
We need to apply the above identity to expand the expression $(-2 x+5 y-3 z)^2$.
$
\begin{aligned}
& (-2 x+5 y-3 z)^2=[(-2 x)+5 y+(-3 z)]^2 \\
& =(-2 x)^2+(5 y)^2+(-3 z)^2+2 \times(-2 x) \times 5 y+2 \times 5 y \times(-3 z)+2 \times(-3 z) \times(-2 x) \\
& =4 x^2+25 y^2+9 z^2-20 x y-30 y z+12 z x .
\end{aligned}
$
(vi) $\left(\frac{1}{4} a-\frac{1}{2} b+1\right)^2$
We know that $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$.
$
\begin{aligned}
& \left(\frac{1}{4} a-\frac{1}{2} b+1\right)^2=\left[\frac{a}{4}+\left(-\frac{b}{2}\right)+1\right]^2 \\
& =\left(\frac{a}{4}\right)^2+\left(-\frac{b}{2}\right)^2+(1)^2+2 \times \frac{a}{4} \times\left(-\frac{b}{2}\right)+2 \times\left(-\frac{b}{2}\right) \times 1+2 \times 1 \times \frac{a}{4} \\
& =\frac{a^2}{16}+\frac{b^2}{4}+1-\frac{a b}{4}-b+\frac{a}{2} .
\end{aligned}
$
Ex 2.4 Question 5.
Factorize:
(i) $4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
(ii) $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
Answer.
(i) $4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
The expression $4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$ can also be written as
$
(2 x)^2+(3 y)^2+(-4 z)^2+2 \times 2 x \times 3 y+2 \times 3 y \times(-4 z)+2 \times(-4 z) \times 2 x
$
We can observe that, we can apply the identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$ with respect to the expression $(2 x)^2+(3 y)^2+(-4 z)^2+2 \times 2 x \times 3 y+2 \times 3 y \times(-4 z)+2 \times(-4 z) \times 2 x$, to get
$
(2 x+3 y-4 z)^2
$
Therefore, we conclude that after factorizing the expression $4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$, we get $(2 x+3 y-4 z)^2$.
(ii) $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
We need to factorize the expression $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$.
The expression $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$ can also be written as
$
(-\sqrt{2} x)^2+(y)^2+(2 \sqrt{2} z)^2+2 \times(-\sqrt{2} x) \times y+2 \times y \times(2 \sqrt{2} z)+2 \times(2 \sqrt{2} z) \times(-\sqrt{2} x) .
$$
We can observe that, we can apply the identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$ with respect to the expression $(-\sqrt{2} x)^2+(y)^2+(2 \sqrt{2} z)^2+2 \times(-\sqrt{2} x) \times y+2 \times y \times(2 \sqrt{2} z)+2 \times(2 \sqrt{2} z) \times(-\sqrt{2} x)$, to get $(-\sqrt{2} x+y+2 \sqrt{2} z)^2$
Therefore, we conclude that after factorizing the expression $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$, we get $(-\sqrt{2} x+y+2 \sqrt{2} z)^2$.
Ex 2.4 Question 6.
Write the following cubes in expanded form:
(i) $(2 x+1)^3$
(ii) $(2 a-3 b)^3$
(iii) $\left(\frac{3}{2} x+1\right)^3$
(iv) $\left(x-\frac{2}{3} y\right)^3$
Answer.
(i) $(2 x+1)^3$
We know that $(x+y)^3=x^3+y^3+3 x y(x+y)$.
$
\begin{aligned}
\therefore(2 x+1)^3 & =(2 x)^3+(1)^3+3 \times 2 x \times 1(2 x+1) \\
& =8 x^3+1+6 x(2 x+1) \\
& =8 x^3+12 x^2+6 x+1 .
\end{aligned}
$
Therefore, the expansion of the expression $(2 x+1)^3$ is $8 x^3+12 x^2+6 x+1$.
(ii) $(2 a-3 b)^3$
We know that $(x-y)^3=x^3-y^3-3 x y(x-y)$.
$
\begin{aligned}
\therefore(2 a-3 b)^3 & =(2 a)^3-(3 b)^3-3 \times 2 a \times 3 b(2 a-3 b) \\
& =8 a^3-27 b^3-18 a b(2 a-3 b) \\
& =8 a^3-36 a^2 b+54 a b^2-27 b^3 .
\end{aligned}
$
Therefore, the expansion of the expression $(2 a-3 b)^3$ is $8 a^3-36 a^2 b+54 a b^2-27 b^3$.
(iii) $\left(\frac{3}{2} x+1\right)^3$
We know that $(x+y)^3=x^3+y^3+3 x y(x+y)$.
$
\begin{aligned}
\left(\frac{3}{2} x+1\right)^3 & =\left(\frac{3}{2} x\right)^3+(1)^3+3 \times \frac{3}{2} x \times 1\left(\frac{3}{2} x+1\right) \therefore \\
& =\frac{27}{8} x^3+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right) \\
& =\frac{27}{8} x^3+\frac{27}{4} x^2+\frac{9}{2} x+1
\end{aligned}
$
Therefore, the expansion of the expression $\left(\frac{3}{2} x+1\right)^3$ is $\frac{27}{8} x^3+\frac{27}{4} x^2+\frac{9}{2} x+1$.
(iv) $\left(x-\frac{2}{3} y\right)^3$
We know that $
$
\begin{aligned}
& \therefore\left(x-\frac{2}{3} y\right)^3=(x)^3-\left(\frac{2}{3} y\right)^3-3 \times x \times \frac{2}{3} y\left(x-\frac{2}{3} y\right) \\
& =x^3-\frac{8}{27} y^3-2 x y\left(x-\frac{2}{3} y\right) \\
& =x^3-2 x^2 y+\frac{4}{3} x y^2-\frac{8}{27} y^3 .
\end{aligned}
$
Therefore, the expansion of the expression $\left(x-\frac{2}{3} y\right)_{\text {is }}^3 x^3-2 x^2 y+\frac{4}{3} x y^2-\frac{8}{27} y^3$.
x-y)^3=x^3-y^3-3 x y(x-y)$.
Ex 2.4 Question 7.
Evaluate the following using suitable identities:
(i) $(99)^3$
(ii) $(102)^3$
(iii) $(998)^3$
Answer.
(i) $(99)^3$
$(99)^3$ can al so be written as $(100-1)^3$.
Using identity, $(x-y)^3=x^3-y^3-3 x y(x-y)$
$
\begin{aligned}
& (100-1)^3=(100)^3-(1)^3-3 \times 100 \times 1(100-1) \\
& =1000000-1-300(99) \\
& =999999-29700 \\
& =970299
\end{aligned}
$
(ii) $(102)^3$
$(102)^3$ can also be written as $(100+2)^3$.
Using identity $(x+y)^3=x^3+y^3+3 x y(x+y)$
$
\begin{aligned}
& (100+2)^3=(100)^3+(2)^3+3 \times 100 \times 2(100+2) \\
& =1000000+8+600(102) \\
& =1000008+61200 \\
& =1061208
\end{aligned}
$
(iii) $(998)^3$
$(998)^3$ can also be written as $(1000-2)^3$.
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$
$
\begin{aligned}
& (1000-2)^3=(1000)^3-(2)^3-3 \times 1000 \times 2(1000-2) \\
& =1000000000-8-6000(998) \\
& =999999992-5988000 \\
& =994011992
\end{aligned}
$
Ex 2.4 Question 8.
Factorize each of the following:
(i) $8 a^3+b^3+12 a^2 b+6 a b^2$ (ii) $8 a^3-b^3-12 a^2 b+6 a b^2$
(iii) $27-125 a^3-135 a+225 a^2$ (iv) $64 a^3-27 b^3-144 a^2 b+108 a b^2$
(v) $27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$
Answer.
(i) $8 a^3+b^3+12 a^2 b+6 a b^2$
The expression $8 a^3+b^3+12 a^2 b+6 a b^2$ can also be written as
$
\begin{aligned}
& =(2 a)^3+(b)^3+3 \times 2 a \times 2 a \times b+3 \times 2 a \times b \times b \\
& =(2 a)^3+(b)^3+3 \times 2 a \times b(2 a+b) .
\end{aligned}
$
Using identity $(x+y)^3=x^3+y^3+3 x y(x+y)$ withrespect to the expression $(2 a)^3+(b)^3+3 \times 2 a \times b(2 a+b)$, we get $(2 a+b)^3$.
Therefore, after factorizing the expression $8 a^3+b^3+12 a^2 b+6 a b^2$, we get $(2 a+b)^3$.
(ii) $8 a^3-b^3-12 a^2 b+6 a b^2$
The expression $8 a^3-b^3-12 a^2 b+6 a b^2$ can also be written as
$
\begin{aligned}
& =(2 a)^3-(b)^3-3 \times 2 a \times 2 a \times b+3 \times 2 a \times b \times b \\
& =(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b) .
\end{aligned}
$
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression $(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b)$ we get $(2 a-b)^3$.
Therefore, after factorizing the expression $8 a^3-b^3-12 a^2 b+6 a b^2$, we get $(2 a-b)^3$.
(iii) $27-125 a^3-135 a+225 a^2$
The expression $27-125 a^3-135 a+225 a^2$ can also be written as
$
\begin{aligned}
& =(3)^3-(5 a)^3-3 \times 3 \times 3 \times 5 a+3 \times 3 \times 5 a \times 5 a \\
& =(3)^3-(5 a)^3+3 \times 3 \times 5 a(3-5 a) .
\end{aligned}
$
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression $(3)^3-(5 a)^3+3 \times 3 \times 5 a(3-5 a)$, we get $(3-5 a)^3$.
Therefore, after factorizing the expression $27-125 a^3-135 a+225 a^2$, we get $(3-5 a)^3$.
(iv) $64 a^3-27 b^3-144 a^2 b+108 a b^2$
The expression $64 a^3-27 b^3-144 a^2 b+108 a b^2$ can also be written as
$
\begin{aligned}
& =(4 a)^3-(3 b)^3-3 \times 4 a \times 4 a \times 3 b+3 \times 4 a \times 3 b \times 3 b \\
& =(4 a)^3-(3 b)^3-3 \times 4 a \times 3 b(4 a-3 b) .
\end{aligned}
$
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression $(4 a)^3-(3 b)^3-3 \times 4 a \times 3 b(4 a-3 b)$ , we get $(4 a-3 b)^3$.
Therefore, after factorizing the expression $64 a^3-27 b^3-144 a^2 b+108 a b^2$, we get $(4 a-3 b)^3$.
(v)
$
27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p
$
The expression $27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$ can also be written as
$
\begin{aligned}
& =(3 p)^3-\left(\frac{1}{6}\right)^3-3 \times 3 p \times 3 p \times \frac{1}{6}+3 \times 3 p \times \frac{1}{6} \times \frac{1}{6} \\
& =(3 p)^3-\left(\frac{1}{6}\right)^3-3 \times 3 p \times \frac{1}{6}\left(3 p-\frac{1}{6}\right)
\end{aligned}
$
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression
$
(3 p)^3-\left(\frac{1}{6}\right)^3-3 \times 3 p \times \frac{1}{6}\left(3 p-\frac{1}{6}\right)
$
to get $\left(3 p-\frac{1}{6}\right)^3$.
$\text { Therefore, after factorizing the expression } 27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p \text {, we get }\left(3 p-\frac{1}{6}\right)^3 \text {. }$
Ex 2.4 Question 9.
Verify:
(i) $x^3+y^3=(x+y)\left(x^2-x y+y^2\right)$
(ii) $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
Answer.
(i) $x^3+y^3=(x+y)\left(x^2-x y+y^2\right)$
We know that $(x+y)^3=x^3+y^3+3 x y(x+y)$.
$
\begin{aligned}
& \Rightarrow x^3+y^3=(x+y)^3-3 x y(x+y) \\
& =(x+y)\left[(x+y)^2-3 x y\right]
\end{aligned}
$
$\because$ We know that $(x+y)^2=x^2+2 x y+y^2$
$
\begin{aligned}
& \therefore x^3+y^3=(x+y)\left(x^2+2 x y+y^2-3 x y\right) \\
& =(x+y)\left(x^2-x y+y^2\right)
\end{aligned}
$
Therefore, the desired result has been verified.
(ii) $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
We know that $(x-y)^3=x^3-y^3-3 x y(x-y)$.
$
\begin{aligned}
& \Rightarrow x^3-y^3=(x-y)^3+3 x y(x-y) \\
& =(x-y)\left[(x-y)^2+3 x y\right]
\end{aligned}
$
$\because$ We know that $(x-y)^2=x^2-2 x y+y^2$
$
\begin{aligned}
& \therefore x^3-y^3=(x-y)\left(x^2-2 x y+y^2+3 x y\right) \\
& =(x-y)\left(x^2+x y+y^2\right)
\end{aligned}
$
Therefore, the desired result has been verified.
Ex 2.4 Question 10.
Factorize:
(i) $27 y^3+125 z^3$
(ii) $64 m^3-343 n^3$
Answer.
(i) $27 y^3+125 z^3$
The expression $27 y^3+125 z^3$ can also be written as $(3 y)^3+(5 z)^3$.
We know that $x^3+y^3=(x+y)\left(x^2-x y+y^2\right)$.
$
\begin{aligned}
& (3 y)^3+(5 z)^3=(3 y+5 z)\left[(3 y)^2-3 y \times 5 z+(5 z)^2\right] \\
& =(3 y+5 z)\left(9 y^2-15 y z+25 z^2\right) .
\end{aligned}
$
(ii) $64 m^3-343 n^3$
The expression $64 m^3-343 n^3$ can also be written as $(4 m)^3-(7 n)^3$.
We know that $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$.
$
\begin{aligned}
(4 m)^3-(7 n)^3 & =(4 m-7 n)\left[(4 m)^2+4 m \times 7 n+(7 n)^2\right] \\
& =(4 m-7 n)\left(16 m^2+28 m n+49 n^2\right)
\end{aligned}
$
Therefore, we conclude that after factorizing the expression $64 m^3-343 n^3$, we get $(4 m-7 n)\left(16 m^2+28 m n+49 n^2\right)$.
Ex 2.4 Question 11.
Factorize: $27 x^3+y^3+z^3-9 x y z$
Answer.
The expression $27 x^3+y^3+z^3-9 x y z$ can also be written as
$
(3 x)^3+(y)^3+(z)^3-3 \times 3 x \times y \times z \text {. }
$
We know that $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$.
$
\begin{aligned}
\therefore(3 x)^3+(y)^3+(z)^3-3 \times 3 x \times y \times z & =(3 x+y+z)\left[(3 x)^2+(y)^2+(z)^2-3 x \times y-y \times z-z \times 3 x\right] \\
& =(3 x+y+z)\left(9 x^2+y^2+z^2-3 x y-y z-3 x z\right) .
\end{aligned}
$
Therefore, we conclude that after factorizing the expression $27 x^3+y^3+z^3-9 x y z$, we get $(3 x+y+z)\left(9 x^2+y^2+z^2-3 x y-y z-3 x z\right)$.
Ex 2.4 Question 12.
Verify that
$
x^3+y^3+z^3-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right]
$
Answer.
LHS is $x^3+y^3+z^3-3 x y z$ and RHS is $\frac{1}{2}(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right]$.
We know that $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$.
And also, we know that $(x-y)^2=x^2-2 x y+y^2$.
$
\begin{aligned}
& \frac{1}{2}(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right] \\
& \frac{1}{2}(x+y+z)\left[\left(x^2-2 x y+y^2\right)+\left(y^2-2 y z+z^2\right)+\left(z^2-2 x z+x^2\right)\right] \\
& \frac{1}{2}(x+y+z)\left(2 x^2+2 y^2+2 z^2-2 x y-2 y z-2 z x\right) \\
& (x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right) .
\end{aligned}
$
Therefore, we can conclude that the desired result is verified.
Ex 2.4 Question 13.
If $x+y+z=0$, show that $x^3+y^3+z^3=0$.
Answer.
We know that $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$.
We need to substitute $x^3+y^3+z^3=0$ in $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$, to get
$
\begin{aligned}
& x^3+y^3+z^3-3 x y=(0)=\left(x^2+y^2+z^2-x y-y z-z x\right), \\
& x^3+y^3+z^3-3 x y z=0 \\
& \Rightarrow x^3+y^3+z^3=3 x y z .
\end{aligned}
$
Therefore, the desired result is verified.
Ex 2.4 Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) $(-12)^3+(7)^3+(5)^3$
(ii) $(28)^3+(-15)^3+(-13)^3$
Answer.
(i) $(-12)^3+(7)^3+(5)^3$
Let $a=-12, b=7$ and $c=5$
We know that, if $a+b+c=0$, then $a^3+b^3+c^3=3 a b c$
Here, $a+b+c=-12+7+5=0$
$
\begin{aligned}
& \therefore(-12)^3+(7)^3+(5)^3=3(-12)(7)(5) \\
& =-1260
\end{aligned}
$
(ii) $(28)^3+(-15)^3+(-13)^3$
Let $a=28, b=-15$ and $c=-13$
We know that, if $a+b+c=0$, then $a^3+b^3+c^3=3 a b c$
Here, $a+b+c=28-15-13=0$
$\begin{aligned}
& \therefore(28)^3+(-15)^3+(-13)^3=3(28)(-15)(-13) \\
& =16380
\end{aligned}$
Ex 2.4 Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Are: $25 a^2-35 a+12$
(ii) Are: $35 y^2+13 y-12$
Answer.
(i) Area : $25 a^2-35 a+12$
The expression $25 a^2-35 a+12$ can also be written as $25 a^2-15 a-20 a+12$.
$
\begin{aligned}
& 25 a^2-15 a-20 a+12=5 a(5 a-3)-4(5 a-3) \\
& =(5 a-4)(5 a-3) .
\end{aligned}
$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $25 a^2-35 a+12$ is Length $=(5 a-4)$ and Breadth $=(5 a-3)$.
(ii) Area : $35 y^2+13 y-12$
The expression $35 y^2+13 y-12$ can also be written as $35 y^2+28 y-15 y-12$.
$
\begin{aligned}
& 35 y^2+28 y-15 y-12=7 y(5 y+4)-3(5 y+4) \\
& =(7 y-3)(5 y+4) .
\end{aligned}
$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $35 y^2+13 y-12$ is Length $=(7 y-3)$ and Breadth $=(5 y+4)$.
Ex 2.4 Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: $3 x^2-12 x$
(ii) Volume: $12 k y^2+8 k y-20 k$
Answer.
(i) Volume : $3 x^2-12 x$
The expression $3 x^2-12 x$ can also be written as $3 \times x \times(x-4)$.
Therefore, we can conclude that a possible expression for the dimension of a cuboid of volume $3 x^2-12 x_{\text {is }} 3, x$ and $(x-4)$.
(ii) Volume : $12 k y^2+8 k y-20 k$
The expression $12 k y^2+8 k y-20 k$ can also be written as $k\left(12 y^2+8 y-20\right)$.
$
\begin{aligned}
& k\left(12 y^2+8 y-20\right)=k\left(12 y^2-12 y+20 y-20\right) \\
& =k[12 y(y-1)+20(y-1)] \\
& =k(12 y+20)(y-1) \\
& =4 k \times(3 y+5) \times(y-1) .
\end{aligned}
$
Therefore, we can conclude that a possible expression for the dimension of a cuboid of volume $12 k y^2+8 k y-20 k_{\text {is }} 4 k,(3 y+5)$ and $(y-1)$.
