NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.5 2026-27

Think and Reflect

James and Reshma were talking about algebraic identities they learnt in school.

James: (a – b)2 (a + b) = (a2 – 2ab + b2)(a + b)

Reshma: I have a different idea: (a – b)2 (a + b) = (a – b) [(a-b) (a + b)] = (a-b)(a2 – b2)

I will find this product to get the answer. According to you, who is correct and why? Try to combine more identities and find new results. 

Solution:

Both methods are correct. They use different algebraic identities, but both lead to the same final polynomial:  (a3 – a2b – ab2 + b3). 

Think and Reflect

We already know that x2 – y2 = (x – y)(x + y)

Further, we have verified that x3 – y3 = (x -y)(x2 + xy + y2) Observe that x – y is a common factor of x2 – y2 and x3 – y3. Do you think x-y is also a factor of x4 – y4?

Note that x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2).

Can you see how x – y is a factor of x4 – y4?

How about x5 – y5? Does this also have x – y as a factor? 

Solution:

Here x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2)

We know that x2 – y2 = (x – y) (x + y)

So, x4 – y4 = (x – y) (x + y) (x2 + y2)

∴ Yes, (x – y) is also a factor of x4 – y4.

Now x3 – y3 = (x – y) (x2 + xy + y2)

x4 – y4 = (x – y) (x + y) (x2 + y2)

= (x – y) x(x2 + y2) + y(x2 + y2)

– (x – y) (x3 + xy2 + x2y + y3)

Now, x5 – y5 = (x – y) (x4 + x3y + xy3 + x2y2 + y4)

∴ Yes, x – y is also a factor of x5 – y5 

Think and Reflect

Try to simplify the following rational expression:

$\dfrac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\dfrac{(6s-t)^2}{(?+?)(?+?)}$

(Hint: Factor t2 + 2ts – 48s2 and simplify the rational expressions assuming that t2 + 2ts – 48s2 * 0).

Solution:

36s2 – 12st + t2 = (6s)2 – 2(6s)t + t2

= (6s – t)2

= t2 + 2ts – 48s2

= t2 + [8s + (-6s)]t + (-6s)(8s)

= [t + 8s] [t + (-6s)]

= (t + 8s) (t – 6s)

= -(t + 8s) (6s – t) 

$\dfrac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\dfrac{(6s-t)^2}{(t+8s)(-6s+t)}=\dfrac{(6s-t)(6s-t)}{-(t+8s)(6s-t)}$

= $\dfrac{-(6s-t)}{(t+8s)}=\dfrac{t+6s}{t+8s}$

= $\dfrac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \dfrac{t + 6s}{t + 8s}$

Exercise Set 4.5 

Question 1.

Simplify the following rational expressions, assuming that the expressions in the denominators are not equal to zero: 

(i) $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

Solution:

$\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

We have,

3p2 – 3pq – 18q2 = 3 (p2 – pq – 6q2)

= 3(p2 – 3pq + 2pq – 6q2)

(∵ -6 = -3 × 2 and -3 + 2 = -1)

= 3[p(p – 3q) + 2q(p – 3q)]

3p2 – 3pq – 18q2 = 3(p + 2q)(p – 3q) …(i)

Now, p2 + 3pq – 10q2 = p2 + 5pq – 2pq – 10q2

[∵ -10 = 5 × (-2) and 5 + (-2) = 3]

= p(p + 5q) – 2q(p + 5q)

Therefore, p2 + 3pq – 10q2 = (p – 2q)(p + 5q) …(ii)

From (i) and (ii), we get 

$\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \dfrac{3(p+2q)(p-3q)}{(p-2q)(p+5q)}$

(ii)  $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

Solution:

Given the expression: $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

Step 1: Factorise the numerator and denominator.

Numerator: The expression n3 – 3n2m + 3nm2 – m3 is a perfect cube expansion and factors as (n – mf

Denominator: The expression 5m2 – 10mn + 5n2 factors as 5(m – n)2

Step 2: Simplify the expression.

Now, the expression becomes $\dfrac{(n-m)^3}{5(m-n)^2}$

Since (n – m) = -(m – n), we get

$\dfrac{-(m-n)^3}{5(m-n)^2}=\dfrac{-(m-n)}{5}=\frac{n-m}{5}$

(iii) \dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}

Solution:

We have, 

$\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

Using the identity a3 + b3 + c3 – 3abc = (a + b + c)

(a2 + b2 + c2 -ab – bc – ca),

take a = w,b = -v, c = x, then

w3 – v3 + x3 + 3wvx = (w – v + x) (w2 + v2 + x2 + wv + vx – wx)

Also, the denominator

w2 + v2 + x2 – 2wv – 2vx + 2wx = (w – v + x)2

Hence, 

=$\dfrac{(w-v+x)(w^2+v^2+x^2+wv+vx-wx)}{(w-v+x)^2}$

= $\dfrac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}$

(iv) $\dfrac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2}$

Solution:

$\dfrac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2}$

4y2 – 20yz + 25z2 = (2y – 5z)2

25z2 – 4y2 = (5z – 2y)(5z + 2y) = $\dfrac{(2y-5z)^2}{(5z-2y)(5z+2y)}$

Since (2y – 5z) = -(5z – 2y), (2y – 5z)2 = (5z – 2y)2

So, $\dfrac{(5z-2y)^2}{(5z-2y)(5z+2y)}=\dfrac{5z-2y}{5z+2y}$

(v) $\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$

Solution: 

We have, 

$\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$

Now, numerator = (x2 + x – 6)(x2 -7x + 12)

= [x2 + 3x – 2x – 6].[x2 – 4x – 3x + 12]

= [x(x + 3) – 2(x + 3)].[x(x – 4) -3(x – 4)]

= [(x + 3)(x – 2)].[(x – 4)(x – 3)]

and denominator = (x2 – 6x + 8)(x2 – 9)

= [x2 – 4x – 2x + 8].(x – 3)(x + 3)

[va2 – b2 = (a + b)(a – b)]

= [x(x – 4) – 2(x – 4)] .(x – 3)(x + 3)

= (x – 4) (x – 2) (x – 3) (x + 3)

Now, 

$\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$

=$\dfrac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} = 1$

(vi) $\dfrac{p^4 - 16}{p^2 - 4p + 4}$

Solution:

$\dfrac{p^4 - 16}{p^2 - 4p + 4} = \dfrac{(p^2+4)(p^2-4)}{(p-2)^2}$

= $\dfrac{(p^2+4)(p-2)(p+2)}{(p-2)^2}$

= $\dfrac{(p^2+4)(p+2)}{p-2}$

Key Benefits of Vedantu’s Class 9 Maths Chapter 4 Exercise 4.5 Solutions

Here are some key benefits of using these solutions:

• Understand the application of identities: The solutions show how algebraic identities simplify expressions and solve different types of questions in Exercise 4.5.

• Learn step-by-step methods: The Class 9 Maths NCERT Solutions Chapter 4 Exercise 4.5 explain each calculation in a logical sequence, making difficult questions easier to follow.

• Improve accuracy: Check your work against the NCERT Solutions Class 9 Maths Chapter 4 Exercise 4.5 to spot calculation errors and see where your approach needs improvement.

• Build independent problem-solving skills: Try each NCERT question yourself first, then refer to the solutions. This helps you learn the method instead of simply copying the final answer.

• Revise important concepts quickly: Use the solutions PDF for last-minute revision of the identities and methods covered in Exercise 4.5.

• Prepare for similar questions: Understanding the steps in the NCERT Maths Class 9 Chapter 4 Exercise 4.5 Solutions helps you apply the same concepts to questions with different expressions or values.

• Support complete Class 9 Maths preparation: After completing Exercise 4.5, you can use the downloaded Class 9 Maths Solutions PDF for exam preparation.

Access Exercise-wise NCERT Solutions for Chapter 4 Maths Class 9

CBSE Class 9 Maths Chapter 4 Exploring Algebraic Identities Study Materials

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