# RD Sharma Class 9 Algebraic Identities Exercise 4.3 Solutions

On this page you will find Maths RD Sharma Class 9 Algebraic Identities Exercise 4.3 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 4.3 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 4.3 below and it will be beneficial for them.

## RD Sharma Class 9 Algebraic Identities Exercise 4.3 Solutions

Q1. Find the cube of each of the following binomial expression
(a) $\left(\frac{1}{x}+\frac{y}{3}\right)$
(b) $\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$
(c) $\left(2 x+\frac{3}{x}\right)$
(d) $\left(4-\frac{1}{3 x}\right)$
Sol:
(a) $\left.\left(\frac{1}{x}+\frac{y}{3}\right)\right)^{3}$
Given, $\left.\left(\frac{1}{x}+\frac{y}{3}\right)\right)^{3}$
The above equation is in the form of $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
We know that, $a=\frac{1}{x}, b=\frac{y}{3}$
By using $(a+b)^{3}$ formula
$\left.\left(\frac{1}{x}+\frac{y}{3}\right)\right)^{3}=\left(\frac{1}{x}\right)^{3}+\left(\frac{y}{3}\right)^{3}+3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+3 * \frac{1}{x} * \frac{y}{3}\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x}\left(\frac{1}{x}+\frac{y}{3}\right)$
$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\left(\frac{y}{x} * \frac{1}{x}\right)+$
$\left(\frac{y}{x} * \frac{y}{3}\right)$
$\left.=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}\right)$
Hence,
$\left.\left.\left(\frac{1}{x}+\frac{y}{3}\right)\right)^{3}=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}\right)$
(b) $\left(\left(\frac{3}{x}-\frac{2}{x^{2}}\right)\right)^{3}$
Given, $\left(\left(\frac{3}{x}-\frac{2}{x^{2}}\right)\right)^{3}$
The above equation is in the form of $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
We know that, $a=\frac{3}{x}, b=\frac{2}{x^{2}}$
By using $(a-b)^{3}$ formula
$\left(\left(\frac{3}{x}-\frac{2}{x^{2}}\right)\right)^{3}=\left(\frac{3}{x}\right)^{3}-\left(\frac{2}{x^{2}}\right)^{3}-3\left(\frac{3}{x}\right)\left(\frac{2}{x^{2}}\right)\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$
$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-3 * \frac{3}{x} * \frac{2}{x^{2}}\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$
$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{18}{x^{3}}\left(\frac{3}{x}-\frac{2}{x^{2}}\right)$
$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\left(\frac{18}{x^{3}} * \frac{3}{x}\right)+\left(\frac{18}{x^{3}} * \frac{2}{x^{2}}\right)$
$=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{54}{x^{4}}+\frac{36}{x^{5}}$
Hence, $\left(\left(\frac{3}{x}-\frac{2}{x^{2}}\right)\right)^{3}=\frac{27}{x^{3}}-\frac{8}{x^{6}}-\frac{54}{x^{4}}+\frac{36}{x^{5}}$
(c) $\left(2 x+\frac{3}{x}\right)^{3}$
Given, $\left(2 x+\frac{3}{x}\right)^{3}$
The above equation is in the form of $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
We know that, $\mathrm{a}=2 \mathrm{x}, \mathrm{b}=\frac{3}{\mathrm{x}}$
By using $(a+b)^{3}$ formula
$=8 x^{3}+\frac{27}{x^{3}}+\frac{18 x}{x}\left(2 x+\frac{3}{x}\right)$
$=8 x^{3}+\frac{27}{x^{3}}+\frac{18 x}{x}\left(2 x+\frac{3}{x}\right)$
$=8 x^{3}+\frac{27}{x^{3}}+$
$(18 * 2 x)+\left(18 * \frac{3}{x}\right)$
$\left.=8 x^{3}+\frac{27}{x^{3}}+36 \times \frac{54}{x}\right)$
Hence,
The cube of $\left.\left(2 x+\frac{3}{x}\right)^{3}=8 x^{3}+\frac{27}{x^{3}}+36 x \frac{54}{x}\right)$
(d) $\left(4-\frac{1}{3 \mathrm{x}}\right)^{3}$
Given, $\left(4-\frac{1}{3 x}\right)^{3}$
The above equation is in the form of $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
We know that, $\mathrm{a}=4, \mathrm{~b}=\frac{1}{3 \mathrm{x}}$
By using $(a-b)^{3}$ formula
$\left(4-\frac{1}{3 x}\right)^{3}=4^{3}-\left(\frac{1}{3 x}\right)^{3}-3(4)\left(\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$
$=64-\frac{1}{27 x^{3}}-\frac{12}{3 x}\left(4-\frac{1}{3 x}\right)$
$=64-\frac{1}{27 x^{3}}-\frac{4}{x}\left(4-\frac{1}{3 x}\right)$
$=64-\frac{1}{27 x^{3}}-\left(\frac{4}{3 x} * 4\right)+\left(\frac{4}{3 x} * \frac{1}{3 x}\right)$
$=64-\frac{1}{27 x^{3}}-\frac{16}{x}+\left(\frac{4}{3 x^{2}}\right.$
Hence,
The cube of $\left(4-\frac{1}{3 x}\right)^{3}=64-\frac{1}{27 x^{3}}-\frac{16}{x}+\left(\frac{4}{3 x^{2}}\right)$

Q2. Simplify each of the following
(a) $(x+3)^{3}+(x-3)^{3}$
(b) $\left(\frac{x}{2}+\frac{y}{3}\right)^{3}-\left(\frac{x}{2}-\frac{y}{3}\right)^{3}$
(c) $\left(x+\frac{2}{x}\right)^{3}+\left(x-\frac{2}{x}\right)^{3}$
(d) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$
Sol:
(a) $(x+3)^{3}+(x-3)^{3}$
The above equation is in the form of $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
We know that, $a=(x+3), b=(x-3)$
By using $\left(a^{3}+b^{3}\right)$ formula
$=(x+3+x-3)\left[(x+3)^{3}+(x-3)^{3}-(x+3)(x-3)\right]$
$=2 x\left[\left(x^{2}+3^{2}+2^{*} x^{\star} 3\right)\right.$ $\left.+\left(x^{2}+3^{2}-2^{*} x^{*} 3\right)-\left(x^{2}-3^{2}\right)\right]$
$=2 x\left[\left(x^{2}+9+6 x\right)+\left(x^{2}+9-6 x\right)-x^{2}+9\right]$
$=2 x\left[\left(x^{2}+9+6 x+x^{2}-9-6 x-x^{2}+9\right)\right]$
$=2 x\left(x^{2}+27\right)$
$=2 x^{3}+54 x$
Hence, the result of $(x+3)^{3}+(x-3)^{3}$ is $2 x^{3}+54 x$
(b) $\left(\frac{x}{2}+\frac{y}{3}\right)^{3}-\left(\frac{x}{2}-\frac{y}{3}\right)^{3}$
The above equation is in the form of $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
We know that, $a=\left(\frac{x}{2}+\frac{y}{3}\right)^{3}, b=\left(\frac{x}{2}-\frac{y}{3}\right)^{3}$
By using $\left(a^{3}-b^{3}\right)$ formula
$=\left[\left(\left(\frac{x}{2}+\frac{y}{3}\right)^{3}-\left(\left(\frac{x}{2}-\frac{y}{3}\right)^{3}\right)\right]\left[\left(\left(\frac{x}{2}+\frac{y}{3}\right)^{3}\right)^{2}\right.\right.$ $\left(\left(\frac{x}{2}-\frac{y}{3}\right)^{3}\right)^{2}-\left(\left(\frac{x}{2}+\frac{y}{3}\right)^{3}\right)\left(\left(\frac{x}{2}-\frac{y}{3}\right)^{3}\right)$
$=\left(\frac{x}{3}+\frac{y}{3}-\frac{x}{2}+\frac{y}{3}\right)\left[\left(\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{3}\right)^{2}+\left(\frac{2 x y}{6}\right)\right)\right.$ $\left.+\left(\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{3}\right)^{2}-\left(\frac{2 x y}{6}\right)\right)+\left(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{3}\right)^{2}\right)\right]$
$=\frac{2 y}{3}\left[\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{2 x y}{6}\right)+\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}-\frac{2 x y}{6}\right)+\frac{x^{2}}{4}-\frac{y^{2}}{9}\right]$
$=\frac{2 y}{3}\left[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{2 x y}{6}+\frac{x^{2}}{4}+\frac{y^{2}}{9}-\frac{2 x y}{6}+\frac{x^{2}}{4}-\frac{y^{2}}{9}\right]$
$=\frac{2 y}{3}\left[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{x^{2}}{4}+\frac{x^{2}}{4}\right]$
$=\frac{2 y}{3}\left[\frac{3 x^{2}}{4}+\frac{y^{2}}{9}\right]$
$=\frac{x^{2} y}{2}+\frac{2 y^{3}}{27}$
Hence, the result of $\left(\frac{x}{2}+\frac{y}{3}\right)^{3}-\left(\frac{x}{2}-\frac{y}{3}\right)^{3}=\frac{x^{2} y}{2}+\frac{2 y^{3}}{27}$
(c) $\left(x+\frac{2}{x}\right)^{3}+\left(x-\frac{2}{x}\right)^{3}$
The above equation is in the form of $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
We know that, $\mathrm{a}=\left(\mathrm{x}+\frac{2}{\mathrm{x}}\right)^{3}, \mathrm{~b}=\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)^{3}$
By using $\left(a^{3}+b^{3}\right)$ formula
$=\left(x+\frac{2}{x}+x-\frac{2}{x}\right)\left[\left(x+\frac{2}{x}\right)^{2}\right.$ $\left.+\left(x-\frac{2}{x}\right)^{2}-\left(\left(x+\frac{2}{x}\right)\left(x-\frac{2}{x}\right)\right)\right]$
$=(2 x)\left[\left(x^{2}+\frac{4}{x^{2}}+\frac{4 x}{x}\right)\right.$ $+\left(x^{2}+\frac{4}{x^{2}}-\frac{4 x}{x}\right)-\left(x^{2}-\frac{4}{x^{2}}\right)$
$=(2 x)\left[\left(x^{2}+\frac{4}{x^{2}}+\frac{4 x}{x}+x^{2}+\frac{4}{x^{2}}-\frac{4 x}{x}-x^{2}+\frac{4}{x^{2}}\right)\right.$
$=(2 x)\left[\left(x^{2}+\frac{4}{x^{2}}+\frac{4}{x^{2}}+\frac{4}{x^{2}}\right)\right.$
$=(2 x)\left[\left(x^{2}+\frac{12}{x^{2}}\right)\right.$
$=2 x^{3}+\frac{24}{x}$
Hence, the result of $\left(x+\frac{2}{x}\right)^{3}+\left(x-\frac{2}{x}\right)^{3}=(2 x)\left[\left(x^{2}+\frac{12}{x^{2}}\right)\right.$
(d) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$
Given, $(2 x-5 y)^{3}-(2 x+5 y)^{3}$
The above equation is in the form of $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
We know that, $a=(2 x-5 y), b=(2 x+5 y)$
By using $\left(a^{3}-b^{3}\right)$ formula
$=(2 x-5 y-2 x-5 y)\left[(2 x-5 y)^{2}\right.$ $\left.+(2 x+5 y)^{2}+((2 x-5 y) *(2 x+5 y))\right]$
$=(-10 y)\left[\left(4 x^{2}+25 y^{2}-20 x y\right)\right.$ $\left.+\left(4 x^{2}+25 y^{2}+20 x y\right)+4 x^{2}-25 y^{2}\right]$
$=(-10 y)\left[4 x^{2}+25 y^{2}-20 x y+4 x^{2}+25 y^{2}+20 x y+4 x^{2}-25 y^{2}\right]$
$=(-10 y)\left[4 x^{2}+4 x^{2}+4 x^{2}+25 y^{2}\right]$
$=(-10 y)\left[12 x^{2}+25 y^{2}\right\}$
$=-120 x^{2} y-250 y^{3}$
Hence, the result of $(2 x-5 y)^{3}-(2 x+5 y)^{3}=-120 x^{2} y-250 y^{3}$

Q3. If $a+b=10$ and $a b=21$, Find the value of $a^{3}+b^{3}$
Sol:
Given,
$a+b=10, a b=21$
we know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ ……..(1)
substitute $a+b=10, a b=21$ in eq 1
$\Rightarrow(10)^{3}=a^{3}+b^{3}+3(21)(10)$
$\Rightarrow 1000=a^{3}+b^{3}+630$
$\Rightarrow 1000-630=a^{3}+b^{3}$
$\Rightarrow 370=a^{3}+b^{3}$
Hence, the value of $\mathrm{a}^{3}+\mathrm{b}^{3}=370$

Q4. If $a-b=4$ and $a b=21$, Find the value of $a^{3}-b^{3}$
Sol:
Given,
$a-b=4, a b=21$
we know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ ………(1)
substitute $a-b=4, a b=21$ in eq 1
$\Rightarrow(4)^{3}=a^{3}-b^{3}-3(21)(4)$
$\Rightarrow 64=a^{3}-b^{3}-252$
$\Rightarrow 64+252=a^{3}-b^{3}$
$\Rightarrow 316=a^{3}-b^{3}$
Hence, the value of $a^{3}-b^{3}=316$

05. If $\left(x+\frac{1}{x}\right)=5$, Find the value of $x^{3}+\frac{1}{x^{3}}$
Sol:
Given, $\left(x+\frac{1}{x}\right)=5$
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ ………(1)
Substitute $\left(x+\frac{1}{x}\right)=5$ in eq 1
$\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x * \frac{1}{x}\right)\left(x+\frac{1}{x}\right)$
$5^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x * \frac{1}{x}\right)\left(x+\frac{1}{x}\right)$
$125=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)$
$125=x^{3}+\frac{1}{x^{3}}+3(5)$
$125=x^{3}+\frac{1}{x^{3}}+15$
$125-15=x^{3}+\frac{1}{x^{3}}$
$x^{3}+\frac{1}{x^{3}}=110$
hence, the result is $x^{3}+\frac{1}{x^{3}}=110$

Q6. If $\left(x-\frac{1}{x}\right)=7$, Find the value of $x^{3}-\frac{1}{x^{3}}$
Sol:
Given, If $\left(x-\frac{1}{x}\right)=7$
We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ ………….(1)
Substitute $\left(x-\frac{1}{x}\right)=7$ in eq 1
$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x * \frac{1}{x}\right)\left(x-\frac{1}{x}\right)$
$7^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$
$343=x^{3}-\frac{1}{x^{3}}-(3 * 7)$
$343=x^{3}-\frac{1}{x^{3}}-(3 * 7)$
$343=x^{3}-\frac{1}{x^{3}}-21$
$343+21=x^{3}-\frac{1}{x^{3}}$
$x^{3}-\frac{1}{x^{3}}=364$
hence, the result is $x^{3}-\frac{1}{x^{3}}=364$

Q7. If $\left(x-\frac{1}{x}\right)=5$, Find the value of $x^{3}-\frac{1}{x^{3}}$
Sol:
Given, If $\left(x-\frac{1}{x}\right)=5$
We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ ……….(1)
Substitute $\left(x-\frac{1}{x}\right)=5$ in eq 1
$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x * \frac{1}{x}\right)\left(x-\frac{1}{x}\right)$
$5^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$
$125=x^{3}-\frac{1}{x^{3}}-(3 * 5)$
$125=x^{3}-\frac{1}{x^{3}}-15$
$125+15=x^{3}-\frac{1}{x^{3}}$
$x^{3}-\frac{1}{x^{3}}=140$
hence, the result is $x^{3}-\frac{1}{x^{3}}=140$

Q8. If $\left(x^{2}+\frac{1}{x^{2}}\right)=51$, Find the value of $x^{3}-\frac{1}{x^{3}}$
Sol:
Given, $\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)=51$
We know that,$(x-y)^{2}=x^{2}+y^{2}-2 x y$ ………….(1)
Substitute $\left(x^{2}+\frac{1}{x^{2}}\right)=51$ in eq 1
$\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-2 * x * \frac{1}{x}$
$\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-2$
$\left(x-\frac{1}{x}\right)^{2}=51-2$
$\left(x-\frac{1}{x}\right)^{2}=49$
$\left(x-\frac{1}{x}\right)=\sqrt{49}$
$\left(x-\frac{1}{x}\right)=\pm 7$
We need to find $\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}$
So, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
$x^{3}-\frac{1}{x^{3}}=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^{2}}+\left(x * \frac{1}{x}\right)\right.$
We know that,
$\left(x-\frac{1}{x}\right)=7$ and $\left(x^{2}+\frac{1}{x^{2}}\right)=51$
$x^{3}-\frac{1}{x^{3}}=7(51+1)$
$x^{3}-\frac{1}{x^{3}}=7(52)$
$x^{3}-\frac{1}{x^{3}}=364$
Hence, the value of $\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}=364$

Q9. If $\left(x^{2}+\frac{1}{x^{2}}\right)=98$, Find the value of $x^{3}+\frac{1}{x^{3}}$
Sol:
Given, $\left(x^{2}+\frac{1}{x^{2}}\right)=98$
We know that,$(x+y)^{2}=x^{2}+y^{2}+2 x y$ ………(1)
Substitute $\left(x^{2}+\frac{1}{x^{2}}\right)=98$ in eq 1
$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2 * x * \frac{1}{x}$
$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$
$\left(x+\frac{1}{x}\right)^{2}=98+2$
$\left(x+\frac{1}{x}\right)^{2}=100$
$\left(x+\frac{1}{x}\right)=\sqrt{100}$
$\left(x+\frac{1}{x}\right)=\pm 10$
We need to find $\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}$
So, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}-\left(x * \frac{1}{x}\right)\right.$
We know that,
$\left(x+\frac{1}{x}\right)=10$ and $\left(x^{2}+\frac{1}{x^{2}}\right)=98$
$x^{3}+\frac{1}{x^{3}}=10(98-1)$
$x^{3}+\frac{1}{x^{3}}=10(97)$
$x^{3}+\frac{1}{x^{3}}=970$
Hence, the value of $\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}=970$

Q10. If $2 x+3 y=13$ and $x y=6$, Find the value of $8 x^{3}+27 y^{3}$
Sol:
Given $, 2 x+3 y=13, x y=6$
We know that,
$(2 x+3 y)^{3}=13^{2}$
$\Rightarrow 8 x^{3}+27 y^{3}+3(2 x)(3 y)(2 x+3 y)=2197$
$\Rightarrow 8 x^{3}+27 y^{3}+18 x y(2 x+3 y)=2197$
Substitute $2 x+3 y=13, x y=6$
$\Rightarrow 8 x^{3}+27 y^{3}+18(6)(13)=2197$
$\Rightarrow 8 x^{3}+27 y^{3}+1404=2197$
$\Rightarrow 8 x^{3}+27 y^{3}=2197-1404$
$\Rightarrow 8 x^{3}+27 y^{3}=793$
Hence, the value of $8 x^{3}+27 y^{3}=793$

Q11. If $3 x-2 y=11$ and $x y=12$, Find the value of $27 x^{3}-8 y^{3}$
Sol:
Given, $3 x-2 y=11, x y=12$
We know that $(a-b)^{3}=a^{3}-b^{3}-3 a b(a+b)$
$(3 x-2 y)^{3}=11^{3}$
$\Rightarrow 27 x^{3}-8 y^{3}-(18 * 12 * 11)=1331$
$\Rightarrow 27 x^{3}-8 y^{3}-2376=1331$
$\Rightarrow 27 x^{3}-8 y^{3}=1331+2376$
$\Rightarrow 27 x^{3}-8 y^{3}=3707$
Hence, the value of $27 x^{3}-8 y^{3}=3707$

Q12. If $\mathrm{x}^{4}+\left(\frac{1}{\mathrm{x}^{4}}\right)=119$, Find the value of $\mathrm{x}^{3}-\left(\frac{1}{\mathrm{x}^{3}}\right)$
Sol:
Given, $\mathrm{x}^{4}+\left(\frac{1}{\mathrm{x}^{4}}\right)=119 \quad—1$
We know that $(x+y)^{2}=x^{2}+y^{2}+2 x y$
Substitute $x^{4}+\left(\frac{1}{x^{4}}\right)=119$ in eq 1
$\left(\mathrm{x}^{2}+\left(\frac{1}{\mathrm{x}^{2}}\right)\right)^{2}=\mathrm{x}^{4}$ $+\left(\frac{1}{x^{4}}\right)+\left(2^{\star} x^{2 *} \frac{1}{x^{2}}\right)$
$=x^{4}+\left(\frac{1}{x^{4}}\right)+2$
$=119+2$
$=121$
$\left(x^{2}+\left(\frac{1}{x^{2}}\right)\right)^{2}=121$
$x^{2}+\left(\frac{1}{x^{2}}\right)=\sqrt{121}$
$x^{2}+\left(\frac{1}{x^{2}}\right)=\pm 11$
Now, find $\left(x-\frac{1}{x}\right)$
We know that $(x-y)^{2}=x^{2}+y^{2}-2 x y$
$\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-\left(2^{*} x^{*} \frac{1}{x}\right.$
$=x^{2}+\frac{1}{x^{2}}-2$
$=11-2$
$=9$
$\left(x-\frac{1}{x}\right)=\sqrt{9}$
$=\pm 3$
We need to find $\mathrm{x}^{3}-\left(\frac{1}{\mathrm{x}^{3}}\right)$
We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}-a b\right)$
$x^{3}-\left(\frac{1}{x^{3}}\right)=\left(x-\frac{1}{x}\right)\left(x^{2}+\left(\frac{1}{x^{2}}\right)+x * \frac{1}{x}\right.$
Here, $\mathrm{x}^{2}+\left(\frac{1}{\mathrm{x}^{2}}\right)=11$ and $\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)=3$
$\mathrm{x}^{3}-\left(\frac{1}{\mathrm{x}^{3}}\right)=3(11+1)$
$=3(12)$
$=36$
Hence, the value of $\mathrm{x}^{3}-\left(\frac{1}{\mathrm{x}^{3}}\right)=36$

Q13. Evaluate each of the following
(a) (103) $^{3}$
(b) $(98)^{3}$
(c) $(9.9)^{3}$
(d) $(10.4)^{3}$
(e) $(598)^{3}$
(f) $(99)^{3}$
Sol:
Given,
(a) $(103)^{3}$
we know that $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$\Rightarrow(103)^{3}$ can be written as $(100+3)^{3}$
Here, $a=100$ and $b=3$
$(103)^{3}=(100+3)^{3}$
$=(100)^{3}+(3)^{3}+3(100)(3)(100+3)$
$=1000000+27+(900 * 103)$
$=1000000+27+92700$
$=1092727$
The value of $(103)^{3}=1092727$
(b) $(98)^{3}$
we know that $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$\Rightarrow(98)^{3}$ can be written as $(100-2)^{3}$
Here, $a=100$ and $b=2$
$(98)^{3}=(100-2)^{3}$
$=(100)^{3}-(2)^{3}-3(100)(2)(100-2)$
$=1000000-8-\left(600^{\star} 102\right)$
$=1000000-8-58800$
$=941192$
The value of $(98)^{3}=941192$
(c) $(9.9)^{3}$
we know that $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$\Rightarrow(9.9)^{3}$ can be written as $(10-0.1)^{3}$
Here, $a=10$ and $b=0.1$
$(9.9)^{3}=(10-0.1)^{3}$
$=(10)^{3}-(0.1)^{3}-3(10)(0.1)(10-0.1)$
$=1000-0.001-(3 * 9.9)$
$=1000-0.001-29.7$
$=1000-29.701$
$=970.299$
The value of $(9.9)^{3}=970.299$
(d) $(10.4)^{3}$
we know that $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$\Rightarrow(10.4)^{3}$ can be written as $(10+0.4)^{3}$
Here, $a=10$ and $b=0.4$
$(10.4)^{3}=(10+0.4)^{3}$
$=(10)^{3}+(0.4)^{3}+3(10)(0.4)(10+0.4)$
$=1000+0.064+(12 * 10.4)$
$=1000+0.064+124.8$
$=1000+124.864$
$=1124.864$
The value of $(10.4)^{3}=1124.864$
(e) $(598)^{3}$
we know that $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$\Rightarrow(598)^{3}$ can be written as $(600-2)^{3}$
Here, $a=600$ and $b=2$
$(598)^{3}=(600-2)^{3}$
$=(600)^{3}-(2)^{3}-3(600)(2)(600-2)$
$=216000000-8-(3600 * 598)$
$=216000000-8-2152800$
$=216000000-2152808$
$=213847192$
The value of $(598)^{3}=213847192$
(f) $(99)^{3}$
we know that $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$\Rightarrow(99)^{3}$ can be written as $(100-1)^{3}$
Here, $a=100$ and $b=1$
$(99)^{3}=(100-1)^{3}$
$=(100)^{3}-(1)^{3}-3(100)(1)(100-1)$
$=1000000-1-\left(300^{\star} 99\right)$
$=1000000-1-29700$
$=1000000-29701$
$=970299$
The value of $(99)^{3}=970299$

Q14. Evaluate each of the following
(a) $111^{3}-89^{3}$
(b) $46^{3}+34^{3}$
(c) $104^{3}+96^{3}$
(d) $93^{3}-107^{3}$
Sol:
Given,
(a) $111^{3}-89^{3}$
the above equation can be written as $(100+11)^{3}-(100-11)^{3}$
we know that, $(a+b)^{3}-(a-b)^{3}=2\left[b^{3}+3 a b^{2}\right]$
here, $\mathrm{a}=100 \mathrm{~b}=11$
$(100+11)^{3}-(100-11)^{3}=2\left[11^{3}+3(100)^{2}(11)\right]$
$=2[1331+330000]$
$=2[331331]$
$=662662$
The value of $111^{3}-89^{3}=662662$
(b) $46^{3}+34^{3}$
the above equation can be written as $(40+6)^{3}+(40-6)^{3}$
we know that, $(a+b)^{3}+(a-b)^{3}=2\left[a^{3}+3 a b^{2}\right]$
here, $a=40, b=4$
$(40+6)^{3}+(40-6)^{3}=2\left[40^{3}+3(6)^{2}(40)\right]$
$=2[64000+4320]$
$=2[68320]$
$=1366340$
The value of $46^{3}+34^{3}=1366340$
(c) $104^{3}+96^{3}$
the above equation can be written as $(100+4)^{3}+(100-4)^{3}$
we know that, $(a+b)^{3}+(a-b)^{3}=2\left[a^{3}+3 a b^{2}\right]$
here, $a=100 b=4$
$(100+4)^{3}-(100-4)^{3}=2\left[100^{3}+3(4)^{2}(100)\right]$
$=2[1000000+4800]$
$=2[1004800]$
$=2009600$
The value of $104^{3}+96^{3}=2009600$
(a) $93^{3}-107^{3}$
the above equation can be written as $(100-7)^{3}-(100+7)^{3}$
we know that, $(a-b)^{3}-(a+b)^{3}=-2\left[b^{3}+3 b a^{2}\right]$
here, $a=93, b=107$
$(100-7)^{3}-(100+7)^{3}=-2\left[7^{3}+3(100)^{2}(7)\right]$
$=-2[343+210000]$
$=-2[210343]$
$=-420686$
The value of $93^{3}-107^{3}=-420686$

Q15. If $x+\frac{1}{x}=3$, calculate $x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}}, x^{4}+\frac{1}{x^{4}}$
Sol:
Given, $x+\frac{1}{x}=3$
We know that $(x+y)^{2}=x^{2}+y^{2}+2 x y$
$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+\left(2 * x * \frac{1}{x}\right)$
$3^{2}=x^{2}+\frac{1}{x^{2}}+2$
$9-2=x^{2}+\frac{1}{x^{2}}$
$x^{2}+\frac{1}{x^{2}}=7$
squaring on both sides
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=7^{2}$
$x^{4}+\frac{1}{x^{4}}+2 * x^{2} * \frac{1}{x^{2}}=49$
$x^{4}+\frac{1}{x^{4}}+2=49$
$x^{4}+\frac{1}{x^{4}} \quad=49-2$
$x^{4}+\frac{1}{x^{4}} \quad=47$
again, cubing on both sides
$\left(x+\frac{1}{x}\right)^{3}=3^{3}$
$x^{3}+\frac{1}{x^{3}}+3 x^{*} \frac{1}{x}\left(x+\frac{1}{x}\right)=27$
$x^{3}+\frac{1}{x^{3}}+\left(3^{*} 3\right)=27$
$x^{3}+\frac{1}{x^{3}}+9=27$
$x^{3}+\frac{1}{x^{3}} \quad=27-9$
$x^{3}+\frac{1}{x^{3}} \quad=18$
hence, the values are $x^{2}+\frac{1}{x^{2}}=7, x^{4}+\frac{1}{x^{4}}=47, x^{3}+\frac{1}{x^{3}}=18$

Q16. If $x^{4}+\frac{1}{x^{4}}=194$, calculate $x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}}, x+\frac{1}{x}$
Sol:
Given,
$x^{4}+\frac{1}{x^{4}}=194 \quad—1$
add and subtract $\left(2^{\star} x^{2} * \frac{1}{x^{2}}\right)$ on left side in above given equation
$x^{4}+\frac{1}{x^{4}}+\left(2^{*} x^{2} * \frac{1}{x^{2}}\right)-2\left(2^{\star} x^{2} * \frac{1}{x^{2}}\right)$ $=194$
$x^{4}+\frac{1}{x^{4}}+\left(2^{*} x^{2} * \frac{1}{x^{2}}\right)-2=194$
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}-2=194$
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=194+2$
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=196$
$\left(x^{2}+\frac{1}{x^{2}}\right)=\sqrt{196}$
$\left(x^{2}+\frac{1}{x^{2}}\right)=14 \quad—-2$
Add and subtract $\left(2^{\star} x^{\star} \frac{1}{x}\right)$ on left side in eq 2
$\left(x^{2}+\frac{1}{x^{2}}\right)+\left(2^{\star} x^{\star} \frac{1}{x}\right)-\left(2^{*} x^{*} \frac{1}{x}\right)=14$
$\left(x+\frac{1}{x}\right)^{2}-2=14$
$\left(x+\frac{1}{x}\right)^{2}=14+2$
$\left(x+\frac{1}{x}\right)^{2}=16$
$\left(x+\frac{1}{x}\right)=\sqrt{16}$
$\left(x+\frac{1}{x}\right)=4 \quad—-3$
Now, cubing eq 3 on both sides
$\left(x+\frac{1}{x}\right)^{3}=4^{3}$
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$x^{3}+\frac{1}{x^{3}}+3^{\star} x^{\star} \frac{1}{x}\left(x+\frac{1}{x}\right)=64$
$x^{3}+\frac{1}{x^{3}}+\left(3^{*} 4\right)=64$
$x^{3}+\frac{1}{x^{3}}=64-12$
$x^{3}+\frac{1}{x^{3}}=52$
hence, the values of $\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=196,\left(x+\frac{1}{x}\right)=4, x^{3}+\frac{1}{x^{3}}=52$

017. Find the values of $27 x^{3}+8 y^{3}$, if
(a) $3 x+2 y=14$ and $x y=8$
(b) $3 x+2 y=20$ and $x y=\frac{14}{9}$
Sol:
(a) Given, $3 x+2 y=14$ and $x y=8$
cubing on both sides
$(3 x+2 y)^{3}=14^{3}$
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$27 x^{3}+8 y^{3}+3(3 x)(2 y)(3 x+2 y)=2744$
$27 x^{3}+8 y^{3}+18 x y(3 x+2 y)=2744$
$27 x^{3}+8 y^{3}+18(8)(14)=2744$
$27 x^{3}+8 y^{3}+2016=2744$
$27 x^{3}+8 y^{3}=2744-2016$
$27 x^{3}+8 y^{3}=728$
Hence, the value of $27 x^{3}+8 y^{3}=728$
(b) Given, $3 x+2 y=20$ and $x y=\frac{14}{9}$
cubing on both sides
$(3 x+2 y)^{3}=20^{3}$
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$27 x^{3}+8 y^{3}+3(3 x)(2 y)(3 x+2 y)=8000$
$27 x^{3}+8 y^{3}+18 x y(3 x+2 y)=8000$
$27 x^{3}+8 y^{3}+18\left(\frac{14}{9}\right)(20)=8000$
$27 x^{3}+8 y^{3}+560=8000$
$27 x^{3}+8 y^{3}=8000-560$
$27 x^{3}+8 y^{3}=7440$
Hence, the value of $27 x^{3}+8 y^{3}=7440$

Q18. Find the value of $64 x^{3}-125 z^{3}$, if $4 x-5 z=16$ and $x z=12$
Sol:
Given, $64 x^{3}-125 z^{3}$
Here, $4 x-5 z=16$ and $x z=12$
Cubing $4 x-5 z=16$ on both sides
$(4 x-5 z)^{3}=16^{3}$
We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$(4 x)^{3}-(5 z)^{3}-3(4 x)(5 z)(4 x-5 z)=16^{3}$
$64 x^{3}-125 z^{3}-60(x z)(16)=4096$
$64 x^{3}-125 z^{3}-60(12)(16)=4096$
$64 x^{3}-125 z^{3}-11520=4096$
$64 x^{3}-125 z^{3}=4096+11520$
$64 x^{3}-125 z^{3}=15616$
The value of $64 x^{3}-125 z^{3}=15616$

Q19. If $x-\frac{1}{x}=3+2 \sqrt{2}$, Find the value of $x^{3}-\frac{1}{x^{3}}$
Sol:
Given, $x-\frac{1}{x}=3+2 \sqrt{2}$
Cubing $x-\frac{1}{x}=3+2 \sqrt{2}$ on both sides
We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$\left(x-\frac{1}{x}\right)^{3}=(3+2 \sqrt{2})^{3}$
$x^{3}-\frac{1}{x^{3}}-3^{\star} x^{*} \frac{1}{x}\left(x-\frac{1}{x}\right)=3^{2}$ $+(2 \sqrt{2})^{3}+3^{*} 3 * 2 \sqrt{2}(3+2 \sqrt{2})$
$x^{3}-\frac{1}{x^{3}}-3(3+2 \sqrt{2})=27+16 \sqrt{2}+18 \sqrt{2}(3+2 \sqrt{2})$
$x^{3}-\frac{1}{x^{3}}=27+16 \sqrt{2}+54 \sqrt{2}+72+9+6 \sqrt{2}$
$x^{3}-\frac{1}{x^{3}}=108+76 \sqrt{2}$
hence, the value of $x^{3}-\frac{1}{x^{3}}=108+76 \sqrt{2}$

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