Class 9 RD Sharma Solutions Chapter 5 Factorisation of Algebraic Expressions Exercise 5.3 (original) (raw)

Last Updated : 23 Jul, 2025

In this article, we will be going to solve the entire exercise 5.3 of RD Sharma's book. Factorization of algebraic expressions is the process of rewriting an expression as a product of its factors. This technique simplifies expressions, solves equations, and helps in various algebraic manipulations.

Exercise 5.3 focuses on factoring algebraic expressions using standard algebraic identities. This skill is crucial for simplifying complex expressions and solving higher-level mathematical problems. The main identities covered are:

1. (a + b)² = a² + 2ab + b²

2. (a - b)² = a² - 2ab + b²

3. a² - b² = (a + b)(a - b)

4. a³ + b³ = (a + b)(a² - ab + b²)

5. a³ - b³ = (a - b)(a² + ab + b²)

What is the Factorisation of Algebraic Expressions?

Factorization of **algebraic expressions involves breaking down a complex expression into simpler factors that, when multiplied together, give the original expression. This process is essential in simplifying and solving equations.

Question 1. Factorize 64a3+125b3+240a2b+300ab2

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(4a)3+(5b)3+3(4a)2(5b)+3(4a)(5b)2

(4a+5b)3

(4a+5b)(4a+5b)(4a+5b)

Hence, 64a3+125b3+240a2b+300ab2 = (4a+5b)(4a+5b)(4a+5b)

Question 2. Factorize 125x3-27y3-225x2y+135xy2

**Solution:

As we know that, a3 - b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(5x)3-(3y)3-3(5x)2(3y)+3(5x)(3y)2

(5x-3y)3

(5x-3y)(5x-3y)(5x-3y)

Hence, 125x3-27y3-225x2y+135xy2 = (5x-3y)(5x-3y)(5x-3y)

Question 3. Factorize 8/27x3+1+4/3x2+2x

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2/3x)3+13+3(2/3x)2(1)+ 3(2/3x)(1)

(2/3x+1)3

(2/3x+1) (2/3x+1) (2/3x+1)

Hence, 8/27x3+1+4/3x2+2x = (2/3x+1) (2/3x+1) (2/3x+1)

Question 4. Factorize 8x3+27y3+36x2y+54xy2

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2x)3+(3y)3+3(2x)2(3y)+3(2x)(3y)2

(2x+3y)3

(2x+3y)(2x+3y)(2x+3y)

Hence, 8x3+27y3+36x2y+54xy2 = (2x+3y)(2x+3y)(2x+3y)

Question 5. Factorize a3-3a2b+3ab2-b3+8

**Solution:

As we know that, a3 - b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(a-b)3 + (2)3

(a-b+2)[(a-b)2-(a-b)(2)+22]

Use a3+b3=(a+b)(a2-ab+b2)

(a-b+2)(a2+b2-2ab-2a+2b+4)

Hence, a3-3a2b+3ab2-b3+8 = (a-b+2)(a2+b2-2ab-2a+2b+4)

Question 6. Factorize x3+8y3+6x2y+12xy2

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

x3+(2y)3+3(x)2(2y)+3(x)(2y)2

(x+2y)3

(x+2y)(x+2y)(x+2y)

Hence, x3+8y3+6x2y+12xy2 = (x+2y)(x+2y)(x+2y)

Question 7. Factorize 8x3+y3+12x2y+6xy2

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2x)3 +y3+3(2x)2y+3(2x)(y)2

(2x+y)3

(2x+y)(2x+y)(2x+y)

Hence, 8x3+y3+12x2y+6xy2 = (2x+y)(2x+y)(2x+y)

Question 8. Factorize 8a3+27b3+36a2b+54ab2

**Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2a)3 +(3b)3+3(2a)2(3b) + 3(2a)(3b)2

(2a+3b)3

(2a+3b) (2a+3b) (2a+3b)

Hence, 8a3+27b3+36a2b+54ab2 = (2a+3b) (2a+3b) (2a+3b)

Question 9. Factorize 8a3-27b3-36a2b+54ab2

**Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(2a)3 -(3b)3-3(2a)2(3b) + 3(2a)(3b)2

(2a-3b)3

(2a-3b) (2a-3b) (2a-3b)

Hence, 8a3-27b3-36a2b+54ab2 = (2a-3b) (2a-3b) (2a-3b)

Question 10. Factorize x3 - 12x(x-4)-64

**Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

x3 -12x2 +48x -43

x3 -43-3(x)2(4) +3(x)(4)2

(x-4)3

(x-4) (x-4) (x-4)

Hence, x3 - 12x(x-4)-64 = (x-4) (x-4) (x-4)

Question 11. Factorize a3x3 -3a2bx2 +3ab2x-b3

**Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(ax)3 -3(ax)2(b)+3(ax)(b)-b3

(ax-b)3

(ax-b) (ax-b) (ax-b)

Hence, a3x3 -3a2bx2 +3ab2x-b3 = (ax-b) (ax-b) (ax-b)

Summary

Exercise 5.3 of RD Sharma's Class 9 Mathematics is a pivotal component in the study of algebraic factorization. This exercise delves into the application of standard algebraic identities to factor expressions efficiently. Students encounter a range of problems involving perfect square trinomials, difference of squares, and sum and difference of cubes. The primary focus is on recognizing these patterns within various algebraic expressions and applying the appropriate identity to simplify them. This skill set is crucial not only for immediate problem-solving but also as a foundation for more advanced mathematical concepts. The exercise emphasizes the importance of pattern recognition and encourages students to verify their factorizations through expansion. By mastering these techniques, students develop a deeper understanding of algebraic structures and relationships. This knowledge proves invaluable in simplifying complex expressions, solving equations, and tackling higher-level mathematical challenges in future coursework. Overall, Exercise 5.3 bridges the gap between basic factorization methods and more sophisticated algebraic manipulations, equipping students with powerful tools for mathematical analysis and problem-solving.