HOTS Questions on Chapter 4 ‘Exploring Algebraic Identities’ for Class 9 with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 4: ‘Exploring Algebraic Identities’ for Class 9 is crafted to push students beyond routine exercises and develop deep, flexible algebraic thinking. Aligned with CBSE and NCERT objectives, the questions emphasise conceptual understanding and creative application of key ideas such as standard identities, factorisation, expansion, verification of identities, and strategic use of identities to simplify expressions and solve problems. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Quick Recap: All Identities from Chapter 4 

Solved HOTS Questions on Algebraic Identities

Question 1: If x + y = 12 and xy = 32, find the value of x² + y².

Solution:

We use the identity:

(x + y)² = x² + 2xy + y²

So: x² + y² = (x + y)² − 2xy

Substituting:

x² + y² = (12)² − 2(32) = 144 − 64 = 80

Question 2: Without multiplying directly, evaluate:

(i) 997² (ii) 1003 × 997

Solution:

(i) Write 997 = (1000 − 3)

(1000 − 3)² = 1000² − 2(1000)(3) + 3²

= 1,000,000 − 6000 + 9 = 994,009

(ii) 1003 × 997 = (1000 + 3)(1000 − 3)

Using (a + b)(a − b) = a² − b²:

= 1000² − 3² = 1,000,000 − 9 = 999,991

Question 3: For any three consecutive square numbers, prove that adding the smallest and largest, then subtracting twice the middle square, always gives 2.

Solution:

Let the three consecutive integers be (n − 1), n, and (n + 1).

Their squares: (n − 1)², n², (n + 1)²

Sum of smallest and largest:

(n − 1)² + (n + 1)²

= (n² − 2n + 1) + (n² + 2n + 1)

= 2n² + 2

Subtracting twice the middle square:

2n² + 2 − 2n² = 2

Since n can be any integer, this is always 2.

Question 4: Find the value of x³ − y³ if x − y = 4 and xy = 5.

Solution: Use x³ − y³ = (x − y)³ + 3xy(x − y)

= (4)³ + 3(5)(4)

= 64 + 60

= 124

Question 5: If a + b + c = 5, ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.

Solution:

We use:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Step 1: Find a² + b² + c²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

25 = a² + b² + c² + 20

a² + b² + c² = 5

Step 2: Find a² + b² + c² − ab − bc − ca

= 5 − 10 = −5

Step 3: Substitute

a³ + b³ + c³ − 3abc = (5)(−5) = −25

Question 6: Check whether n³ − n is always divisible by 6 for all natural numbers n.

Solution:

Factorise: n³ − n = n(n² − 1) = n(n − 1)(n + 1)

This is the product of three consecutive integers: (n − 1), n, (n + 1).

Among any three consecutive integers:

At least one is divisible by 2

Exactly one is divisible by 3

Therefore, their product is divisible by both 2 and 3, hence divisible by 6.

Question 7: Factorise 27u³ − 1/125 − 27u²/5 + 9u/25

Solution:

Rewrite as:

(3u)³ − (1/5)³ − 3(3u)²(1/5) + 3(3u)(1/5)²

= (3u − 1/5)³

Using (a − b)³ = a³ − 3a²b + 3ab² − b³

Answer: (3u − 1/5)³

Question 8: Simplify (x² − 7x + 12) / (5x² + 5x − 100)

Solution:

Numerator: x² − 7x + 12

Find a, b such that a + b = −7 and ab = 12 → a = −3, b = −4

= (x − 3)(x − 4)

Denominator: 5x² + 5x − 100 = 5(x² + x − 20)

Find a, b such that a + b = 1 and ab = −20 → a = 5, b = −4

= 5(x + 5)(x − 4)

Simplify:

= (x − 3)(x − 4) / [5(x + 5)(x − 4)]

= (x − 3) / [5(x + 5)] (cancelling (x − 4), valid since denominator ≠ 0)

Question 9: If x + y = −4, find the value of x³ + y³ − 12xy + 64.

Solution:

Since x + y = −4, we have x + y + 4 = 0.

Let a = x, b = y, c = 4. Then a + b + c = 0.

When a + b + c = 0, the identity gives:

a³ + b³ + c³ = 3abc

So: x³ + y³ + 64 = 3(x)(y)(4) = 12xy

Therefore: x³ + y³ − 12xy + 64 = 12xy − 12xy = 0

Question 10: The volume of a cube is given as p³ + 6p²q + 12pq² + 8q³ cubic units. Find the side length.

Solution:

Compare with the identity (a + b)³ = a³ + 3a²b + 3ab² + b³:

p³ + 6p²q + 12pq² + 8q³

= p³ + 3(p²)(2q) + 3(p)(2q)² + (2q)³

= (p + 2q)³

So the side of the cube = (p + 2q) units

Questions PDF with worked-out examples for Class 9 Chapter 4: Exploring Algebraic Identities, perfect for last-minute CBSE exam revision.

Class 9 Chapter 4: Exploring Algebraic Identities HOTS PDF