RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.3 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.3 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 5 for class 9 deals with the topic of triangles and it is one of the most important chapters.

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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 5 Factorisation of Algebraic Expressions Exercise 5.3 below and it will be beneficial for them.

RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.3 Solutions

Q1. $64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$
SOLUTION :
$=(4 a)^{3}+(5 b)^{3}+3(4 a)^{2}(5 b)+3(4 a)(5 b)^{2}$
$\left[\because a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$
$=(4 a+5 b)^{3}$
$=(4 a+5 b)(4 a+5 b)(4 a+5 b)$
$\therefore 64 \mathrm{a}^{3}+125 \mathrm{~b}^{3}+240 \mathrm{a}^{2} \mathrm{~b}+300 \mathrm{ab}^{2}$ $=(4 a+5 b)(4 a+5 b)(4 a+5 b)$

Q2. $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$
SOLUTION :
$=(5 x)^{3}-(3 y)^{3}-3(5 x)^{2}(3 y)+3(5 x)(3 y)^{2}$ $\left[\because \mathrm{a}^{3}-\mathrm{b}^{3}-3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}-\mathrm{b})^{3}\right]$
$=(5 x-3 y)^{3}$
$=(5 x-3 y)(5 x-3 y)(5 x-3 y)$
$\therefore 125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}=(5 x-3 y)(5 x-3 y)(5 x-3 y)$

Q3. $\frac{8}{27} \mathrm{x}^{3}+1+\frac{4}{3} \mathrm{x}^{2}+2 \mathrm{x}$
SOLUTION :
$=\left(\frac{2}{3} x^{3}\right)^{3}+1^{3}+3 \times\left(\frac{2}{3} x\right)^{2} \times 1+3(1)^{2} \times\left(\frac{2}{3} x\right)$
$=\left(\frac{2}{3} x+1\right)^{3}$ $\left[\because x^{3}+b^{3}+3 x^{2} b+3 x b^{2}=(x+b)^{3}\right]$
$=\left(\frac{2}{3} \mathrm{x}+1\right)\left(\frac{2}{3} \mathrm{x}+1\right)\left(\frac{2}{3} \mathrm{x}+1\right)$
$\therefore \frac{8}{27} \mathrm{x}^{3}+1+\frac{4}{3} \mathrm{x}^{2}+2 \mathrm{x}=\left(\frac{2}{3} \mathrm{x}+1\right)\left(\frac{2}{3} \mathrm{x}+1\right)\left(\frac{2}{3} \mathrm{x}+1\right)$

Q4. $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$
SOLUTION :
$=(2 x)^{3}+(3 y)^{3}+3 \times(2 x)^{2} \times 3 y+3 \times(2 x)(3 y)^{2}$
$=(2 x+3 y)^{3}$ $[\because\left.a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
$\therefore 8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}=(2 x+3 y)(2 x+3 y)(2 x+3 y)$

Q5. $a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$
SOLUTION :
$=(a-b)^{3}+2^{3}$ $\left[\because \mathrm{a}^{3}-\mathrm{b}^{3}-3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}-\mathrm{b})^{3}\right]$
$=(a-b+2)\left((a-b)^{2}-(a-b) 2+2^{2}\right)$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2(a-b)+4\right)$
$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$
$\therefore \mathrm{a}^{3}-3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}-\mathrm{b}^{3}+8=(\mathrm{a}-\mathrm{b}+2)\left(\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab}-2 \mathrm{a}+2 \mathrm{~b}+4\right)$

Q6. $x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}$
SOLUTION :
$=(x)^{3}+(2 y)^{3}+3 \times x^{2} \times 2 y+3 \times x \times(2 y)^{2}$
$=(\mathrm{x}+2 \mathrm{y})^{3}$ $\left[\because x^{3}+y^{3}+3 x^{2} y+3 x y^{2}=(x+y)^{3}\right]$
$=(x+2 y)(x+2 y)(x+2 y)$
$\therefore x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}=(x+2 y)(x+2 y)(x+2 y)$

Q7 $8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}$
SOLUTION :
$=(2 x)^{3}+(y)^{3}+3 \times(2 x)^{2} \times y+3(2 x) \times y^{2}$
$=(2 x+y)^{3}$ $\left[\because \mathrm{a}^{3}+\mathrm{b}^{3}+3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}+\mathrm{b})^{3}\right]$
$=(2 x+y)(2 x+y)(2 x+y)$
$\therefore 8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}=(2 x+y)(2 x+y)(2 x+y)$

Q8. $8 a^{3}+27 b^{3}+36 a^{2} b+54 a b^{2}$
SOLUTION :
$=(2 a)^{3}+(3 b)^{3}+3 \times(2 a)^{2} \times 3 b+3 \times 2 a \times(3 b)^{2}$
$=(2 a+3 b)^{3}$ $\left[\because \mathrm{a}^{3}+\mathrm{b}^{3}+3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}+\mathrm{b})^{3}\right]$
$=(2 a+3 b)(2 a+3 b)(2 a+3 b)$
$\therefore 8 \mathrm{a}^{3}+27 \mathrm{~b}^{3}+36 \mathrm{a}^{2} \mathrm{~b}+54 \mathrm{ab}^{2}=(2 \mathrm{a}+3 \mathrm{~b})(2 \mathrm{a}+3 \mathrm{~b})(2 \mathrm{a}+3 \mathrm{~b})$

Q9. $8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$
SOLUTION :
$=(2 a)^{3}-(3 b)^{3}-3 \times(2 a)^{2} \times 3 b+3 \times 2 a \times(3 b)^{2}$
$=(2 a-3 b)^{3}$ $\left[\because \mathrm{a}^{3}-\mathrm{b}^{3}-3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}-\mathrm{b})^{3}\right]$
$=(2 a-3 b)(2 a-3 b)(2 a-3 b)$
$\therefore 8 \mathrm{a}^{3}-27 \mathrm{~b}^{3}-36 \mathrm{a}^{2} \mathrm{~b}+54 \mathrm{ab}^{2}=(2 \mathrm{a}-3 \mathrm{~b})(2 \mathrm{a}-3 \mathrm{~b})(2 \mathrm{a}-3 \mathrm{~b})$

Q 10 . $x^{3}-12 x(x-4)-64$
SOLUTION :
$=x^{3}-12 x^{2}+48 x-64$
$=x^{3}-3 \times x^{2} \times 4+3 \times 4^{2} \times x-4^{3}$
$=(x-4)^{3}$ $\left[\because a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$
$=(x-4)(x-4)(x-4)$
$\therefore x^{3}-12 x(x-4)-64=(x-4)(x-4)(x-4)$

Q 11 . $a^{3} x^{3}-3 a^{2} b x^{2}+3 a b^{2} x-b^{3}$
SOLUTION :
$=(a x)^{3}-3(a x)^{2} \times b+3(a x) \times b^{2}-b^{3}$
$=(a x-b)^{3}$ $\left[\because \mathrm{a}^{3}-\mathrm{b}^{3}-3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}=(\mathrm{a}-\mathrm{b})^{3}\right]$
$=(a x-b)(a x-b)(a x-b)$
$\therefore a^{3} x^{3}-3 a^{2} b x^{2}+3 a b^{2} x-b^{3}=(a x-b)(a x-b)(a x-b)$

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