# RD Sharma Class 9 Algebraic Identities Exercise 4.4 Solutions

On this page you will find Maths RD Sharma Class 9 Algebraic Identities Exercise 4.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 4 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Algebraic Identities Exercise 4.4  Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 4 Algebraic Identities Exercise 4.4 below and it will be beneficial for them.

## RD Sharma Class 9 Algebraic Identities Exercise 4.4 Solutions

Q1. Find the following products
(a) $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$
(b) $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$
(c) $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$
(d) $\left(\frac{x}{2}+2 y\right)$ $(\left.\frac{x^{2}}{4}-x y+4 y^{2}\right)$
(e) $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
(f) $\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{x^{2}}\right)$
(g) $\left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right)$
(h) $\left(\frac{3}{x}-2 x^{2}\right)\left(\frac{9}{x^{2}}+4 x^{4}-6 x\right)$
(i) $(1-x)\left(1+x+x^2\right)$
(j) $(1+x)\left(1-x+x^2\right)$
(k) $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$
(I) $\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$
Sol:
(a) $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$
Given, $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)$ can we written as
$\left.\Rightarrow(3 x+2 y)\left[(3 x)^{2}-(3 x)(2 y)+(2 y)^{2}\right)\right]$
$\Rightarrow(3 x)^{3}+(2 y)^{3}$
$\Rightarrow 27 x^{3}+8 y^{3}$
Hence, the value of $(3 x+2 y)\left(9 x^{2}-6 x y+4 y^{2}\right)=27 x^{3}+8 y^{3}$
(b) $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$
Given, $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$ can we written as
$\left.\Rightarrow(4 x-5 y)\left[(4 x)^{2}+(4 x)(5 y)+(5 y)^{2}\right)\right]$
$\Rightarrow(4 x)^{3}-(5 y)^{3}$
$\Rightarrow 16 x^{3}-25 y^{3}$
Hence, the value of $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)=16 x^{3}-25 y^{3}$
(c) $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$
Given, $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$ can be written as
$\left.\Rightarrow\left(7 p^{4}+q\right)\left[\left(7 p^{4}\right)^{2}-\left(7 p^{4}\right)(q)+(q)^{2}\right)\right]$
$\Rightarrow\left(7 p^{4}\right)^{3}+(q)^{3}$
$\Rightarrow 343 p^{12}+q^{3}$
Hence, the value of $\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)=343 p^{12}+q^{3}$
(d) $\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$
Sol:
Given, $\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$ can be written as
$\Rightarrow\left(\frac{x}{2}+2 y\right)\left[\left(\frac{x}{2}\right)^{2}-\frac{x}{2}(2 y)+(2 y)^{2}\right]$
$\Rightarrow\left(\frac{x}{2}\right)^{3}+(2 y)^{3}$
$\Rightarrow>\frac{x^{3}}{8}+8 y^{3}$
(e) $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
Sol:
Given, $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
Can be written as,
$\Rightarrow\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{3}{x}\right)^{2}+\left(\frac{5}{y}\right)^{2}+\left(\frac{3}{x}\right)\left(\frac{5}{y}\right)$
$\Rightarrow\left(\frac{3}{x}\right)^{3}-\left(\frac{5}{y}\right)^{3}$
$\Rightarrow\left(\frac{27}{x^{3}}\right)-\left(\frac{125}{y^{3}}\right)$
Hence, the value of $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)=\left(\frac{27}{x^{3}}\right)-\left(\frac{125}{y^{3}}\right)$
(f) $\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{x^{2}}\right)$
Sol:
Given, $\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{x^{2}}\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{x^{2}}\right)$ can be written as,
$\Rightarrow\left(3+\frac{5}{x}\right)\left[\left(3^{2}\right)-3\left(\frac{5}{x}\right)+\left(\frac{5}{x}\right)^{2}\right]$
$\Rightarrow(3)^{3}+\left(\frac{5}{x}\right)^{3}$
$\Rightarrow 27+\frac{125}{x^{3}}$
Hence, the value of $\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{x^{2}}\right)$ is $27+\frac{125}{x^{3}}$
(g) $\left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right)$
Sol:
Given, $\left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right)$ can be written as,
$\Rightarrow\left(\frac{2}{x}+3 x\right)\left[\left(\frac{2}{x}\right)^{2}+(3 x)^{2}-\left(\frac{2}{x}\right)(3 x)\right]$
$\Rightarrow\left(\frac{2}{x}\right)^{3}+(3 x)^{3}$
$\Rightarrow \frac{8}{x^{3}}+9 x^{3}$
Hence, the value of $\left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right)$ is $\frac{8}{x^{3}}+9 x^{3}$
(h) $\left(\frac{3}{x}-2 x^{2}\right)\left(\frac{9}{x^{2}}+4 x^{4}-6 x\right)$
Sol:
Given, $\left(\frac{3}{x}-2 x^{2}\right)\left(\frac{9}{x^{2}}+4 x^{4}-6 x\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\frac{3}{x}-2 x^{2}\right)\left(\frac{9}{x^{2}}+4 x^{4}-6 x\right)$ can be written as,
$\Rightarrow\left(\frac{3}{x}-2 x^{2}\right)\left[\left(\frac{3}{x}\right)^{2}+\left(2 x^{2}\right)^{2}-\left(\frac{3}{x}\right)\left(2 x^{2}\right)\right]$
$\Rightarrow\left(\frac{3}{x}-2 x^{2}\right)\left[\left(\frac{9}{x^{2}}\right)+4 x^{4}-\left(\frac{3}{x}\right)\left(2 x^{2}\right)\right]$
$\Rightarrow\left(\frac{3}{x}\right)^{3}-\left(2 x^{2}\right)^{3}$
$\Rightarrow \frac{27}{x^{3}}-8 x^{6}$
Hence, $\left(\frac{3}{x}-2 x^{2}\right)\left(\frac{9}{x^{2}}+4 x^{4}-6 x\right)$ is $\frac{27}{x^{3}}-8 x^{6}$
(i) $(1-x)\left(1+x+x^2\right)$
Sol:
Given, $(1-x)\left(1+x+x^2\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\Rightarrow(1-x)\left[\left(1^{2}+(1)(x)+x^{2}\right)\right]$
$\Rightarrow(1)^{3}-(x)^{3}$
$\Rightarrow 1-x^{3}$
Hence, the value of $(1-x)\left(1+x+x^2\right)$ is $1-x^{3}$
(j) $(1+x)\left(1-x+x^2\right)$
Sol:
Given, $(1+x)\left(1-x+x^2\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$(1+x)\left(1-x+x^2\right)$ can be written as,
$\Rightarrow(1+x)\left[\left(1^{2}-(1)(x)+x^{2}\right)\right]$
$\Rightarrow(1)^{3}+(x)^{3}$
$\Rightarrow 1+x^{3}$
Hence, the value of $(1+x)\left(1+x-x^2\right)$ is $1+x^{3}$
(k) $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$
Sol:
Given, $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{4}+\mathrm{x}^{2}+1\right)$ can be written as,
$\Rightarrow\left(x^{2}-1\right)\left[\left(x^{2}\right)^{2}-1^{2}+\left(x^{2}\right)(1)\right]$
$\Rightarrow\left(x^{2}\right)^{3}-1^{3}$
$\Rightarrow x^{6}-1$
Hence, $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$ is $x^{6}-1$
(I) $\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$
Sol:
Given, $\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{6}-\mathrm{x}^{3}+1\right)$ can be written as,
$\Rightarrow\left(x^{3}+1\right)\left[\left(x^{3}\right)^{2}-\left(x^{3}\right)(1)+1^{2}\right]$
$\Rightarrow\left(x^{3}\right)^{3}+1^{3}$
$\Rightarrow x^{9}+1$
Hence, the value of $\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$ is $x^{9}+1$

Q2. Find $x=3$ and $y=-1$, Find the values of each of the following using in identity:
(a) $\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$
(b) $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
(c) $\left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right)$
(d) $\left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{x y}{21}\right)$
(e) $\left(\frac{5}{x}+5 x\right)\left(\frac{25}{x^{2}}-25+25 x^{2}\right)$
Sol:
(a) $\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$ can be written as,
$\Rightarrow\left(9 x^{2}-4 x^{2}\right)\left[\left(9 y^{2}\right)^{2}+(9)(4) x^{2} y^{2}+\left(4 x^{2}\right)^{2}\right]$
$\Rightarrow\left(9 y^{2}\right)^{3}-\left(4 x^{2}\right)^{3}$
$\Rightarrow 729 y^{6}-64 x^{6}$
Substitute the value $x=3, y=-1$ in $729 y^{6}-64 x^{6}$ we get,
$\Rightarrow 729 y^{6}-64 x^{6}$
$\Rightarrow 729(-1)^{6}-64(3)^{6}$
$\Rightarrow 729(1)-64(729)$
$\Rightarrow 729-46656$
$\Rightarrow>-45927$
Hence, the product value of $\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)=-45927$
(b) $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
Sol:
Given,
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$
Can be written as,
$\Rightarrow\left(\frac{3}{x}-\frac{x}{3}\right)\left[\left(\frac{3}{x}\right)^{2}+\left(\frac{x}{3}\right)^{2}+\left(\frac{3}{x}\right)\left(\frac{x}{3}\right)\right]$
$\Rightarrow\left(\frac{3}{x}\right)^{3}-\left(\frac{x}{3}\right)^{3}$
$\Rightarrow\left(\frac{27}{x^{3}}\right)-\left(\frac{x^{3}}{27}\right)–1$
Substitute $x=3$ in eq 1
$\Rightarrow\left(\frac{27}{3^{3}}\right)-\left(\frac{3^{3}}{27}\right)$
$\Rightarrow\left(\frac{27}{27}\right)-\left(\frac{27}{27}\right)$
$\Rightarrow 0$
Hence, the value of $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{x^{2}}+\frac{25}{y^{2}}+\frac{15}{x y}\right)$ is 0
(c) $\left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right)$
Sol:
Given,
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right)$
Can be written as,
$\Rightarrow\left(\frac{x}{7}+\frac{y}{3}\right)\left[\left(\frac{x}{7}\right)^{2}+\left(\frac{y}{3}\right)^{2}-\left(\frac{x}{7}\right)\left(\frac{y}{3}\right)\right]$
$\Rightarrow\left(\frac{x}{7}\right)^{3}+\left(\frac{y}{3}\right)^{3}$
$\Rightarrow\left(\frac{x^{3}}{343}\right)+\left(\frac{y^{3}}{27}\right)–1$
Substitute $x=3, y=-1$ in eq 1
$\Rightarrow\left(\frac{3^{3}}{343}\right)+\left(\frac{(-1)^{3}}{27}\right)$
$\Rightarrow\left(\frac{27}{343}\right)-\left(\frac{1}{27}\right)$
Taking least common multiple, we get
$\Rightarrow \frac{27 * 27}{343 * 27}-\frac{1 * 343}{27 * 343}$
$\Rightarrow \frac{729}{9261}-\frac{343}{9261}$
$\Rightarrow \frac{729-343}{9261}$
$\Rightarrow \frac{386}{9261}$
Hence, the value of $\left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right)=\frac{386}{9261}$
(d) $\left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{y^{2}}{9}-+\frac{x y}{21}\right)$
Sol:
Given,
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{x y}{21}\right)$
Can be written as,
$\Rightarrow\left(\frac{x}{4}-\frac{y}{3}\right)\left[\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{3}\right)^{2}+\left(\frac{x}{4}\right)\left(\frac{y}{3}\right)\right]$
$\Rightarrow\left(\frac{x}{4}\right)^{3}-\left(\frac{y}{3}\right)^{3}$
$\Rightarrow\left(\frac{x^{3}}{64}\right)-\left(\frac{y^{3}}{27}\right)–1$
Substitute $x=3, y=-1$ in eq 1
$\Rightarrow\left(\frac{3^{3}}{343}\right)-\left(\frac{(-1)^{3}}{27}\right)$
$\Rightarrow\left(\frac{27}{64}\right)+\left(\frac{1}{27}\right)$
Taking least common multiple, we get
$\Rightarrow \frac{27 * 27}{64 * 27}+\frac{1 * 64}{27 * 64}$
$\Rightarrow \frac{729}{1728}+\frac{64}{1728}$
$\Rightarrow \frac{729+64}{1728}$
$\Rightarrow \frac{793}{9261}$
Hence, the value of $\left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{x y}{21}\right)=\frac{793}{1728}$
(e) $\left(\frac{5}{x}+5 x\right)\left(\frac{25}{x^{2}}-25+25 x^{2}\right)$
Sol:
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(\frac{5}{x}+5 x\right)\left(\frac{25}{x^{2}}-25+25 x^{2}\right)$ can be written as,
$\Rightarrow\left(\frac{5}{x}+5 x\right)\left[\left(\frac{5}{x}\right)^{2}+(5 x)^{2}-\left(\frac{5}{x}\right)(5 x)\right]$
$\Rightarrow\left(\frac{5}{x}\right)^{3}+(5 x)^{3}$
$\Rightarrow \frac{125}{x^{3}}+125 x^{3}—1$
Substitute $x=3$, in eq 1
$\Rightarrow \frac{125}{3^{3}}+125(3)^{3}$
$\Rightarrow \frac{125}{27}+125 * 27$
$\Rightarrow \frac{125}{27}+3375$
Taking least common multiple, we get
$\Rightarrow \frac{125}{27}+\frac{3375 * 27}{27 * 1}$
$\Rightarrow \frac{125}{27}+\frac{91125}{27}$
$=>\frac{125+91125}{27}$
$\Rightarrow \frac{91250}{25}$
Hence, the value of $\left(\frac{5}{x}+5 x\right)\left(\frac{25}{x^{2}}-25+25 x^{2}\right)$ is $\frac{91250}{25}$

Q3. If $a+b=10$ and $a b=16$, find the value of $a^{2}-a b+b^{2}$ and $a^{2}+a b+b^{2}$
Sol:
Given, $a+b=10, a b=16$
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$\Rightarrow a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$
$\Rightarrow a^{3}+b^{3}=(10)^{3}-3(16)(10)$
$\Rightarrow a^{3}+b^{3}=1000-480$
$\Rightarrow a^{3}+b^{3}=520$
Substitute $, a^{3}+b^{3}=520, a+b=10$ in $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$520=10\left(a^{2}+b^{2}-a b\right)$
$\frac{520}{10}=\left(a^{2}+b^{2}-a b\right)$
$\Rightarrow\left(a^{2}+b^{2}-a b\right)=52$
Now, we need to find $\left(a^{2}+b^{2}+a b\right)$
Add and subtract $2 \mathrm{ab}$ in $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{ab}$
$\Rightarrow a^{2}+b^{2}+a b-2 a b+2 a b$
$\Rightarrow(a+b)^{2}-a b$
Substitute $a+b=10, a b$
$\Rightarrow a^{2}+b^{2}+a b=10^{2}-16$
$=100-16$
$=84$
Hence, the values of $\left(a^{2}+b^{2}-a b\right)=52$ and $\left(a^{2}+b^{2}+a b\right)=84$

04. If $a+b=8$ and $a b=6$, find the value of $a^{3}+b^{3}$
Sol:
Given, $a+b=8$ and $a b=6$
We know that, $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$
$\Rightarrow a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$
$\Rightarrow a^{3}+b^{3}=(8)^{3}-3(6)(8)$
$\Rightarrow a^{3}+b^{3}=512-144$
$\Rightarrow a^{3}+b^{3}=368$
Hence, the value of $a^{3}+b^{3}$ is 368

Q5. If $a-b=6$ and $a b=20$, find the value of $a^{3}-b^{3}$
Sol:
Given, $a-b=6$ and $a b=20$
We know that, $a^{3}-b^{3}=(a-b)^{3}+3 a b(a-b)$
$\Rightarrow a^{3}-b^{3}=(a-b)^{3}+3 a b(a-b)$
$\Rightarrow a^{3}-b^{3}=(6)^{3}+3(20)(6)$
$\Rightarrow a^{3}-b^{3}=216+360$
$\Rightarrow a^{3}-b^{3}=576$
Hence, the value of $a^{3}-b^{3}$ is 576

Q6. If $x=-2$ and $y=1$, by using an identity find the value of the following:
(a) $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$
(b) $\left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1\right)$
(c) $\left(5 y+\frac{15}{y}\right)\left(25 y^{2}-75+\frac{225}{y^{2}}\right)$
Sol:
Given,
(a) $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$ can be written as,
$\Rightarrow\left(4 y^{2}-9 x^{2}\right)\left[(4 x)^{2}+4 y^{2 *} 9 x^{2}+\left(9 x^{2}\right)^{2}\right)$
$\Rightarrow\left(4 y^{2}\right)^{3}-\left(9 x^{2}\right)^{3}$
$\Rightarrow 64 y^{6}-729 x^{6}—-1$
Substitute $x=-2$ and $y=1$ in eq 1
$\Rightarrow 64 y^{6}-729 x^{6}$
$\Rightarrow 64(1)^{6}-729(-2)^{6}$
$\Rightarrow 64-729(64)$
$\Rightarrow 64(1-729)$
$\Rightarrow 64(-728)$
$\Rightarrow-46592$
Hence, the value of $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$ is $-46592$
(b) $\left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1\right)$
here $x=-2$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$\left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1\right)$ can be witten as,
$\Rightarrow\left(\frac{2}{x}-\frac{x}{2}\right)\left[\left(\frac{2}{x}\right)^{2}+\left(\frac{x}{2}\right)^{2}+\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)\right]$
$\Rightarrow\left(\frac{2}{x}\right)^{3}-\left(\frac{x}{2}\right)^{3}$
$\Rightarrow\left(\frac{8}{x^{3}}\right)-\left(\frac{x^{3}}{8}\right)—1$
Substitute $x=-2$ in eq 1
$\Rightarrow\left(\frac{8}{(-2)^{3}}\right)-\left(\frac{(-2)^{3}}{8}\right)$
$\Rightarrow\left(\frac{8}{-8}\right)-\left(\frac{-8}{8}\right)$
$\Rightarrow-1+1$
$\Rightarrow 0$
Hence, the value of $\left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1\right)$ is 0
(c) $\left(5 y+\frac{15}{y}\right)\left(25 y^{2}-75+\frac{225}{y^{2}}\right)$
Sol:
We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$\left(5 \mathrm{y}+\frac{15}{\mathrm{y}}\right)\left(25 \mathrm{y}^{2}-75+\frac{225}{\mathrm{y}^{2}}\right)$ can be written as,
$\Rightarrow\left(5 y+\frac{15}{y}\right)\left[(5 y)^{2}+\left(\frac{15}{y}\right)^{2}-(5 y)\left(\frac{15}{y}\right)\right]$
$\Rightarrow(5 y)^{3}+\left(\frac{15}{y}\right)^{3}$
$\Rightarrow 125 y^{3}+\left(\frac{3375}{y^{3}}\right)—1$
Substitute $y=1$ in eq 1
$\Rightarrow 125(1)^{3}+\left(\frac{3375}{(1)^{2}}\right)$
$\Rightarrow 125+3375$
$\Rightarrow 3500$
Hence, the value of $\left(5 \mathrm{y}+\frac{15}{y}\right)\left(25 \mathrm{y}^{2}-75+\frac{225}{\mathrm{y}^{2}}\right)$ is 3500 .

EXERCISE $4.5$

Q1. Find the following products:
(a) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$
(b) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$
(c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$
(d) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$
Sol:
Given,
(a) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$
we know that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
so,
$(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$ $=(3 x)^{3}+(2 y)^{3}+(2 z)^{3}-3(3 x)(2 y)(2 z)$ $=27 x^{3}+8 y^{3}+8 z^{3}-36 x y z$
Hence, the value of $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$ is $27 x^{3}+8 y^{3}+8 z^{3}-36 x y z$
(b) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$
we know that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
SO,
$(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$ $=(4 x)^{3}+(-3 y)^{3}+(2 z)^{3}-3(4 x)(-3 y)(2 z)$ $=64 x^{3}-27 y^{3}+8 z^{3}+72 x y z$
Hence, the value of $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$ is $64 x^{3}-27 y^{3}+8 z^{3}+72 x y z$
(c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$
we know that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
so,
$(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$ $=(2 a)^{3}+(-3 b)^{3}+(-2 c)^{3}-3(2 a)(-3 b)(-2 c)$ $=8 a^{3}-27 b^{3}-8 c^{3}-36 a b c$
Hence, the value of (c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$ is $8 a^{3}-27 b^{3}-8 c^{3}-36 a b c$
(d) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$
we know that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
so,
$(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$ $=(3 x)^{3}+(-4 y)^{3}+(5 z)^{3}-3(3 x)(-4 y)(5 z)$ $=27 x^{3}-64 y^{3}+125 z^{3}+180 x y z$
Hence, the value of $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$ is $27 x^{3}-64 y^{3}+125 z^{3}+180 x y z$

Q2. If $x+y+z=8$ and $x y+y z+z x=20$, Find the value of $x^{3}+y^{3}+z^{3}-3 x y z$
Sol:
given, $x+y+z=8$ and $x y+y z+z x=20$
We know that,
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)$
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(20)$
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+40$
$8^{2}=x^{2}+y^{2}+z^{2}+40$
$64-40=x^{2}+y^{2}+z^{2}$
$x^{2}+y^{2}+z^{2}=24$
we know that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]$
here, $x+y+z=8, x y+y z+z x=20, x^{2}+y^{2}+z^{2}=24$
$x^{3}+y^{3}+z^{3}-3 x y z=8[(24-20)]$
$=8 * 4$
$=32$
Hence, the value of $x^{3}+y^{3}+z^{3}-3 x y z$ is 32

Q3. If $a+b+c=9$ and $a b+b c+c a=26$, Find the value of $a^{3}+b^{3}+c^{3}-3 a b c$
Sol:
Given, $a+b+c=9$ and $a b+b c+c a=26$
We know that,
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(26)$
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+52$
$9^{2}=a^{2}+b^{2}+c^{2}+52$
$81-52=a^{2}+b^{2}+c^{2}$
$a^{2}+b^{2}+c^{2}=29$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a)\right]$
here, $a+b+c=9, a b+b c+c a=26, a^{2}+b^{2}+c^{2}=29$
$a^{3}+b^{3}+c^{3}-3 a b c=9[(29-26)]$
$=9 * 3$
$=27$
Hence, the value of $a^{3}+b^{3}+c^{3}-3 a b c$ is 27

Q4. If $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$, Find the value of $a^{3}+b^{3}+c^{3}-3 a b c$
Sol:
Given, $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$
We know that,
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$9^{2}=35+2(a b+b c+c a)$
$81=35+2(a b+b c+c a)$
$81-35=2(a b+b c+c a)$
$\frac{46}{2}=a b+b c+c a$
$a b+b c+c a=23$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a)\right]$
here, $a+b+c=9, a b+b c+c a=23, a^{2}+b^{2}+c^{2}=35$
$a^{3}+b^{3}+c^{3}-3 a b c=9[(35-23)]$
$=9 * 12$
$=108$
Hence, the value of $a^{3}+b^{3}+c^{3}-3 a b c$ is 108

Q5. Evaluate:
(a) $25^{3}-75^{3}+50^{3}$
(b) $48^{3}-30^{3}-18^{3}$
(c) $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$
(d) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
Sol:
Given,
(a) $25^{3}-75^{3}+50^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
here, $a=25, b=-75, c=50$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(25-75+50)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$25^{3}+(-75)^{3}+50^{3}=3 a b c$
$=3(25)(-75)(50)$
$=-281250$
Hence, the value $25^{3}+(-75)^{3}+50^{3}=-281250$
(b) $48^{3}-30^{3}-18^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
here, $a=48, b=-30, c=-18$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(48-30-18)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$48^{3}+(-30)^{3}+(-18)^{3}=3 a b c$
$=3(48)(-30)(-18)$
$=77760$
Hence, the value $48^{3}+(-30)^{3}+(-18)^{3}=77760$
(c) $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
here, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{1}{3}, \mathrm{c}=\frac{-5}{6}$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
by using least common multiple
$a^{3}+b^{3}+c^{3}=\left(\frac{1 * 6}{2 * 6}+\frac{1 * 4}{3 * 4}-\frac{5 * 2}{6 * 2}\right)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=\left(\frac{6}{12}+\frac{4}{12}-\frac{10}{12}\right)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=0\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{-5}{6}\right)^{3}=3^{*} \frac{1}{2} * \frac{1}{3} * \frac{-5}{6}$
$=\frac{1}{2} * \frac{-5}{6}$
$=\frac{-5}{12}$
Hence, the value of $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$ is $\frac{-5}{12}$
(d) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
here, $a=0.2, b=0.3, c=0.1$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0.2-0.3+0.1)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3 a b c$
$=3(0.2)(-0.3)(0.1)$
$=-0.018$
Hence, the value $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$ is $0.018$

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