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📚 Class IX Maths 🧩 Worksheet Chapter 4: Exploring Algebraic Identities

Algebraic Identities Class 9 Worksheet with Answers PDF

Class 9 Maths Algebraic Identities worksheet with answers — 75 questions with step-by-step solutions. Ganita Manjari Ch 4. Free PDF & online practice.

This free Worksheet for CBSE Class IX Maths, Chapter 4: Exploring Algebraic Identities, contains a structured worksheet with MCQs, short answer, case-based and HOTS questions in one place. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

📌 How to use this Worksheet

"Exploring Algebraic Identities" — Class 09
UNIQUE STUDY POINT BY SUMEET SAHU

"Exploring Algebraic Identities"

Class 09 · Maths (Ganita Manjari) · Practice Worksheet with Solutions

75 Questions
Tap any question's "Show Answer" button to reveal the full step-by-step solution.

Section A · MCQ / Assertion–Reason / True-False (Q1–Q28)

1 Mark each
1
A gardener is designing a rectangular flower bed. The length is given by (3w + 2) meters and the width by (w + 6) meters. What polynomial represents the area of the flower bed?MCQ
  • a) 3w² + 18w + 12 square meters
  • b) 3w² + 20w + 12 square meters
  • c) 3w² + 8w + 12 square meters
  • d) 3w² + 12 square meters
✓ Correct Answer: (b) 3w² + 20w + 12 square meters
  1. Area = length × width = (3w + 2)(w + 6).
  2. = 3w(w + 6) + 2(w + 6) = 3w² + 18w + 2w + 12.
  3. = 3w² + 20w + 12 square meters.
2
If x = a + b + c3, then (x − a)³ + (x − b)³ + (x − c)³ can be factorized as:MCQ
  • a) 3(x−a)(x−b)(x−c)2
  • b) 3(x−a)(x−b)(x−c)
  • c) (x−a)(x−b)(x−c)3
  • d) (x−a)(x−b)(x−c)
✓ Correct Answer: (b) 3(x−a)(x−b)(x−c)
  1. Let p = x−a, q = x−b, r = x−c.
  2. p + q + r = 3x − (a+b+c) = 3·[a+b+c3] − (a+b+c) = (a+b+c) − (a+b+c) = 0.
  3. When p + q + r = 0, the identity p³ + q³ + r³ = 3pqr applies.
  4. So (x−a)³ + (x−b)³ + (x−c)³ = 3(x−a)(x−b)(x−c).
3
If (3m + 2)(9m² − 6m + 4) is expanded, what is the resulting expression?MCQ
  • a) 27m³ − 8
  • b) (3m + 2)³
  • c) 27m³ + 8
  • d) 9m² + 4
✓ Correct Answer: (c) 27m³ + 8
  1. This product matches the identity (x + y)(x² − xy + y²) = x³ + y³.
  2. Here x = 3m and y = 2: x² = 9m², xy = 6m, y² = 4 — matching (9m² − 6m + 4).
  3. So (3m + 2)(9m² − 6m + 4) = (3m)³ + (2)³ = 27m³ + 8.
4
(104 × 96) = ?MCQ
  • a) 9984
  • b) 9684
  • c) 9894
  • d) 9884
✓ Correct Answer: (a) 9984
  1. 104 × 96 = (100 + 4)(100 − 4).
  2. = (100)² − (4)² = 10000 − 16 = 9984.
5
Determine the factors for the expression x² + 14x + 45.MCQ
  • a) (x + 3)(x + 15)
  • b) (x + 1)(x + 45)
  • c) (x + 5)(x + 9)
  • d) (x + 6)(x + 8)
✓ Correct Answer: (c) (x + 5)(x + 9)
  1. We need two numbers whose product is 45 and sum is 14 — these are 5 and 9.
  2. x² + 14x + 45 = x² + 5x + 9x + 45 = x(x + 5) + 9(x + 5) = (x + 5)(x + 9).
6
2a² − 18 in the form of factors is:MCQ
  • a) 2(a − 1)(a − 9)
  • b) 2(a + 3)(a − 3)
  • c) 2(a + 1)(a − 9)
  • d) (2a − 1)(a − 9)
✓ Correct Answer: (b) 2(a + 3)(a − 3)
  1. 2a² − 18 = 2(a² − 9) = 2[(a)² − (3)²].
  2. = 2(a + 3)(a − 3).
7
x⁴ − y⁴ = ________ (x − y)(x² + y²)MCQ
  • a) 0
  • b) x + y
  • c) x − y
  • d) 1
✓ Correct Answer: (b) x + y
  1. x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²).
  2. x² − y² further factors as (x + y)(x − y).
  3. So x⁴ − y⁴ = (x + y)(x − y)(x² + y²), meaning the missing factor is (x + y).
8
The value of [(2.3)³ − 0.027] ÷ [(2.3)² + 0.69 + 0.09] is:MCQ
  • a) 2
  • b) 2.273
  • c) 3
  • d) 2.327
✓ Correct Answer: (a) 2
  1. Write 0.027 = (0.3)³, 0.69 = 2 × 2.3 × 0.3, and 0.09 = (0.3)².
  2. So the expression matches the form (a³ − b³) ÷ (a² + ab + b²) with a = 2.3, b = 0.3.
  3. Since a³ − b³ = (a − b)(a² + ab + b²), the (a² + ab + b²) terms cancel (both non-zero).
  4. Result = a − b = 2.3 − 0.3 = 2.
9
(x² + 3x) men can do a piece of work in (x² − 2x) days, then one day’s work of 1 man is:MCQ
  • a) None of these
  • b) x−2x+3
  • c) x+3x−2
  • d) (x²+3x)(x²−2x)
✓ Correct Answer: (a) None of these
  1. Total men = x² + 3x, total days = x² − 2x, so total work = (x² + 3x)(x² − 2x) man-days.
  2. One day’s work of 1 man = 1 ÷ [(x² + 3x)(x² − 2x)], which doesn’t match any of the simplified options given.
  3. So the answer is “None of these”.
10
The expression m² − 12m + 36 is a perfect square trinomial. Which of the following is its factored form?MCQ
  • a) (m + 12)(m − 3)
  • b) (m + 6)²
  • c) (m − 12)(m + 3)
  • d) (m − 6)²
✓ Correct Answer: (d) (m − 6)²
  1. m² − 12m + 36 matches a² − 2ab + b² with a = m, b = 6.
  2. So it factors to (m − 6)².
11
3ab − 6b + 4a² − 8a in the form of factors is:MCQ
  • a) (a − 2)(4a + 3b)
  • b) (a − 2)(4a − 3b)
  • c) (a + 2)(4a + 3b)
  • d) (a + 2)(4a − 3b)
✓ Correct Answer: (a) (a − 2)(4a + 3b)
  1. 3ab − 6b + 4a² − 8a = 3b(a − 2) + 4a(a − 2).
  2. = (a − 2)(3b + 4a) = (a − 2)(4a + 3b).
12
Factorisation of 6xy − 4y + 6 − 9x isMCQ
  • a) (3y − 2)(2x − 3)
  • b) (3x − 2)(2y − 3)
  • c) (3y − 2)(3x − 2)
  • d) (2y − 3)(2 − 3x)
✓ Correct Answer: (b) (3x − 2)(2y − 3)
  1. Rearrange: 6xy − 9x − 4y + 6.
  2. = 3x(2y − 3) − 2(2y − 3) = (3x − 2)(2y − 3).
13
If x¹⁄³ + y¹⁄³ + z¹⁄³ = 0, thenMCQ
  • a) (x + y + z)³ = 27xyz
  • b) x³ + y³ + z³ = 0
  • c) x³ + y³ + z³ = 27xyz
  • d) x + y + z = 3xyz
✓ Correct Answer: (a) (x + y + z)³ = 27xyz
  1. Let a = x¹⁄³, b = y¹⁄³, c = z¹⁄³, so a + b + c = 0.
  2. Then a³ + b³ + c³ = 3abc, i.e. x + y + z = 3(xyz)¹⁄³.
  3. Cubing both sides: (x + y + z)³ = 3³(xyz) = 27xyz.
14
12a² − 27b⁴MCQ
  • a) 3(2a + 3b²)(2a − b²)
  • b) (2a + 3b²)(2a − 3b²)
  • c) 3(2a + 3b²)(a − 3b²)
  • d) 3(2a + 3b²)(2a − 3b²)
✓ Correct Answer: (d) 3(2a + 3b²)(2a − 3b²)
  1. 12a² − 27b⁴ = 3(4a² − 9b⁴) = 3[(2a)² − (3b²)²].
  2. = 3(2a + 3b²)(2a − 3b²).
15
If x³ − 1x³ = 14, then x − 1x =MCQ
  • a) 5
  • b) 2
  • c) 3
  • d) 4
✓ Correct Answer: (b) 2
  1. Let A = x − 1x. Using a³ − b³ = (a−b)(a²+ab+b²) = (a−b)[(a−b)² + 3ab]:
  2. 14 = A(A² + 3), i.e. A³ + 3A − 14 = 0.
  3. Factoring: A³ − 2A² + 2A² − 4A + 7A − 14 = A²(A−2) + 2A(A−2) + 7(A−2) = (A−2)(A²+2A+7) = 0.
  4. Since A²+2A+7 has no real roots, A − 2 = 0 ⇒ A = 2.
16
The identity (a + b)² = a² + 2ab + b² is known as the:MCQ
  • a) Square of a binomial difference identity
  • b) Square of a binomial sum identity
  • c) Difference of squares identity
  • d) Cubic sum identity
✓ Correct Answer: (b) Square of a binomial sum identity
  1. This identity expands the square of the sum of two terms (a + b), so it is called the square of a binomial sum identity.
17
What is the result of expanding (p + q)(p² − pq + q²)?MCQ
  • a) p³ + q³
  • b) p² + q²
  • c) p³ − q³
  • d) (p + q)³
✓ Correct Answer: (a) p³ + q³
  1. This matches the identity (x + y)(x² − xy + y²) = x³ + y³, with x = p, y = q.
  2. So (p + q)(p² − pq + q²) = p³ + q³.
18
Simplify the expression (a + 2b + c)(a² + 4b² + c² − 2ab − ac − 2bc).MCQ
  • a) a³ + 8b³ + c³ + 6abc
  • b) a³ + 4b³ + c³ − 3abc
  • c) a³ + 8b³ + c³ − 6abc
  • d) a³ − 8b³ + c³ − 6abc
✓ Correct Answer: (c) a³ + 8b³ + c³ − 6abc
  1. This uses the identity (x+y+z)(x²+y²+z²−xy−yz−zx) = x³+y³+z³−3xyz, with x=a, y=2b, z=c.
  2. Checking terms: x²=a², y²=4b², z²=c², −xy=−2ab, −yz=−2bc, −zx=−ac — all match.
  3. Result = a³ + (2b)³ + c³ − 3(a)(2b)(c) = a³ + 8b³ + c³ − 6abc.
19
If x² + 1x² = 102, then x − 1x =MCQ
  • a) 10
  • b) 12
  • c) 13
  • d) 8
✓ Correct Answer: (a) 10
  1. x² + 1x² − 2·x·1x = 102 − 2 ⇒ (x − 1x)² = 100.
  2. x − 1x = √100 = 10.
20
Simplify the rational expression a³−b³a²+ab+b², assuming that a²+ab+b² ≠ 0.MCQ
  • a) a² − b²
  • b) a − b
  • c) a² + b²
  • d) a + b
✓ Correct Answer: (b) a − b
  1. Using a³ − b³ = (a − b)(a² + ab + b²):
  2. a³−b³a²+ab+b² = (a−b)(a²+ab+b²)a²+ab+b² = a − b (cancelling the common non-zero factor).
21
Statement I: [(a²−b²)³+(b²−c²)³+(c²−a²)³] ÷ [(a+b)³+(b+c)³+(c+a)³] = (a+c)(c+a).
Statement II: a²+b²+c² − ab − bc − ca = ½[(a−b)²+(b−c)²+(c−a)²]Statement-Based
  • a) Both Statement-I and Statement-II are true.
  • b) Statement-I is true but Statement-II is false.
  • c) Statement-I is false but Statement-II is true.
  • d) Both Statement-I and Statement-II are false.
✓ Correct Answer: (c) Statement-I is false but Statement-II is true.
  1. Statement I: Since (a²−b²)+(b²−c²)+(c²−a²) = 0, we get (a²−b²)³+(b²−c²)³+(c²−a²)³ = 3(a²−b²)(b²−c²)(c²−a²), and similarly the denominator = 3(a+b)(b+c)(c+a).
  2. The ratio simplifies to (a−b)(a+c)(b−a) … which is not equal to (a+c)(c+a) in general — so Statement I is false.
  3. Statement II: ½[(a−b)²+(b−c)²+(c−a)²] = ½[2a²+2b²+2c²−2ab−2bc−2ca] = a²+b²+c²−ab−bc−ca. This is a standard true identity, so Statement II is true.
22
Assertion (A): To simplify (2p − 3q + 4r)², one must first expand (2p − 3q)² and then treat 4r as a separate term applying (A + B)² form.
Reason (R): The algebraic identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx requires all terms x, y, z to be positive.Assertion-Reason
  • a) Both A and R are true, and R is the correct explanation of A
  • b) Both A and R are true, but R is not the correct explanation of A
  • c) A is true, but R is false
  • d) A is false, but R is true
✓ Correct Answer: (c) A is true, but R is false
  1. Assertion: Grouping (2p−3q) as one term and 4r as another, then applying (A+B)² twice, is indeed a valid way to expand the trinomial square — so A is true.
  2. Reason: The identity (x+y+z)² = x²+y²+z²+2xy+2yz+2zx holds for any real x, y, z, including negative ones (signs are simply carried into the variables) — so R is false.
23
Assertion (A): The expression (K + L − M)² will have three negative terms in its expansion using the general identity for three terms.
Reason (R): The square of a negative term is always positive, but the product of a positive and a negative term is negative.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (d) A is false but R is true.
  1. Assertion: (K+L−M)² = K²+L²+M²+2KL−2LM−2KM, which has only two negative terms (−2LM and −2KM), not three — so A is false.
  2. Reason: The square of any real number is non-negative, and a positive × negative product is indeed negative — this statement is true in general.
24
Assertion (A): Factorisation of x³ − 27 is (x − 3)(x² + 3x + 9).
Reason (R): a³ − b³ = (a − b)(a² + ab + b²).Assertion-Reason
  • a) Both A and R are true, and R is the correct explanation of A
  • b) Both A and R are true, but R is not the correct explanation of A
  • c) A is true, but R is false
  • d) A is false, but R is true
✓ Correct Answer: (a) Both A and R are true, and R is the correct explanation of A
  1. Reason: a³−b³=(a−b)(a²+ab+b²) is a standard true identity.
  2. Assertion: x³ − 27 = x³ − 3³ = (x−3)(x²+3x+3²) = (x−3)(x²+3x+9) — true, and it follows directly by applying Reason’s identity.
25
Assertion (A): The equation x² − 4 = (x − 2)(x + 2) is an algebraic identity.
Reason (R): This equation is true for all real values of x, as expanding the right side yields x² − 4.Assertion-Reason
  • a) Both A and R are true, and R is the correct explanation of A
  • b) Both A and R are true, but R is not the correct explanation of A
  • c) A is true, but R is false
  • d) A is false, but R is true
✓ Correct Answer: (a) Both A and R are true, and R is the correct explanation of A
  1. An algebraic identity is an equality that holds for all values of the variable.
  2. Expanding: (x−2)(x+2) = x(x+2) − 2(x+2) = x²+2x−2x−4 = x²−4, true for every real x.
  3. So Assertion is true, and Reason correctly explains it.
26
Assertion (A): An algebraic identity is a type of equation that holds true for all possible real values of its variables.
Reason (R): The equation (x − y)² = x² − y² is an example of an algebraic identity.Assertion-Reason
  • a) Both A and R are true, and R is the correct explanation of A
  • b) Both A and R are true, but R is not the correct explanation of A
  • c) A is true, but R is false
  • d) A is false, but R is true
✓ Correct Answer: (c) A is true, but R is false
  1. Assertion correctly defines an algebraic identity — true.
  2. Reason: (x−y)² = x² − 2xy + y², which is not the same as x²−y² in general. E.g. x=1, y=2: (1−2)²=1 but 1²−2²=−3. So R is false.
27
State whether the given statement is True or False:
(a) The degree of the polynomial g(x) = (p − x)³ + 14 is 3.
  1. Expanding (p − x)³ gives a term in −x³, so the highest power of x in g(x) is 3.
  2. Adding the constant 14 does not change the highest power.
  3. So the degree of g(x) is indeed 3 — the statement is True.
✓ (a) True
28
Fill in the blanks:
(a) The value of 5.63 × 5.63 + 11.26 × 2.37 + 2.37 × 2.37 is ________.
  1. Notice 11.26 = 2 × 5.63, so the expression is 5.63² + 2(5.63)(2.37) + 2.37².
  2. This matches the identity a² + 2ab + b² = (a + b)² with a = 5.63, b = 2.37.
  3. = (5.63 + 2.37)² = (8)² = 64.
✓ (a) 64

Section B · Match & Short Answer (Q29–Q45)

2 Marks each
29
If x + y = 9 and xy = 20, then find the value of x² + y².
  1. Using the identity (x + y)² = x² + y² + 2xy:
  2. 9² = x² + y² + 2(20) ⇒ 81 = x² + y² + 40.
  3. x² + y² = 81 − 40 = 41.
✓ x² + y² = 41
30
Match the column: (polynomial with its zeroes)
(a) 4x² − 9?
(b) x² − 16?
(c) x² − 8?
(d) t² − 3?
Options: (i) 2√2, −2√2   (ii) √3, −√3   (iii) 32, −32   (iv) 4, −4
  1. (a) 4x² − 9 = 0 ⇒ x² = 94 ⇒ x = ±32 → (iii).
  2. (b) x² − 16 = 0 ⇒ x = ±4 → (iv).
  3. (c) x² − 8 = 0 ⇒ x² = 8 ⇒ x = ±2√2 → (i).
  4. (d) t² − 3 = 0 ⇒ t = ±√3 → (ii).
✓ (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
31
Factor 9x² + 24xy + 16y² completely.
  1. We use the identity a² + 2ab + b² = (a + b)².
  2. 9x² + 24xy + 16y² = (3x)² + 2(3x)(4y) + (4y)².
✓ 9x² + 24xy + 16y² = (3x + 4y)²
32
Simplify: (a + b + c)² + (a − b + c)²
  1. (x+y+z)² = x²+y²+z²+2xy+2yz+2zx and (x−y+z)² = x²+y²+z²−2xy−2yz+2zx, with x=a, y=b, z=c.
  2. (a+b+c)² = a²+b²+c²+2ab+2ac+2bc.
  3. (a−b+c)² = a²+b²+c²−2ab+2ac−2bc.
  4. Adding: the +2ab and −2ab cancel, and +2bc and −2bc cancel, leaving 2a²+2b²+2c²+4ac.
✓ (a+b+c)²+(a−b+c)² = 2a² + 2b² + 2c² + 4ac
33
Expand: (4 − 13x)³
  1. Using (a − b)³ = a³ − b³ − 3ab(a − b), with a = 4, b = 13x:
  2. (4 − 13x)³ = 4³ − 13x³ − 3(4)13x(4 − 13x).
  3. = 64 − 127x³ − 4x(4 − 13x) = 64 − 127x³ − 16x + 43x².
✓ (4 − 13x)³ = 64 − 127x³ − 16x + 43x²
34
If √m + √n − √p = 0, then find the value of (m + n − p)².
  1. From √m + √n − √p = 0, we get √m + √n = √p.
  2. Squaring: (√m + √n)² = (√p)² ⇒ m + n + 2√m√n = p ⇒ m + n − p = −2√mn.
  3. Squaring again: (m + n − p)² = 4mn.
✓ (m + n − p)² = 4mn
35
Factorise: a³ − 8b³ − 64c³ − 24abc
  1. Write as a³ + (−2b)³ + (−4c)³ − 3(a)(−2b)(−4c).
  2. Using a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca), with the three terms a, −2b, −4c:
  3. = (a − 2b − 4c)[a² + 4b² + 16c² − a(−2b) − (−2b)(−4c) − (−4c)(a)]
  4. = (a − 2b − 4c)(a² + 4b² + 16c² + 2ab − 8bc + 4ca).
✓ a³−8b³−64c³−24abc = (a−2b−4c)(a²+4b²+16c²+2ab−8bc+4ca)
36
Factorize: 32a³ + 108b³
  1. 32a³ + 108b³ = 4(8a³ + 27b³) = 4[(2a)³ + (3b)³].
  2. Using a³+b³ = (a+b)(a²−ab+b²): = 4(2a+3b)[(2a)² − (2a)(3b) + (3b)²].
✓ 32a³ + 108b³ = 4(2a + 3b)(4a² − 6ab + 9b²)
37
Factorise the following:
i. a³ + ab(1 − 2a) − 2b²
ii. x² + 1x² − 2 − 3x + 3x
  1. (i) a³ + ab(1−2a) − 2b² = a³ + ab − 2a²b − 2b² = a³ − 2a²b + ab − 2b².
  2. Group: a²(a − 2b) + b(a − 2b) = (a − 2b)(a² + b).
  3. (ii) x² + 1x² − 2 − 3x + 3x = (x² − 2 + 1x²) − 3(x − 1x) = (x − 1x)² − 3(x − 1x).
  4. Let A = x − 1x: expression = A² − 3A = A(A − 3) = (x − 1x)(x − 1x − 3).
✓ (i) (a−2b)(a²+b); (ii) (x − 1x)(x − 1x − 3)
38
Factorise: 8a³ + b³ + 12a²b + 6ab²
  1. 8a³ + b³ + 12a²b + 6ab² = (2a)³ + (b)³ + 3(2a)(b)(2a + b).
  2. This matches (a+b)³ = a³+b³+3ab(a+b), with a → 2a, b → b.
✓ 8a³ + b³ + 12a²b + 6ab² = (2a + b)³
39
Factorize: 27y³ + 125z³
  1. 27y³ + 125z³ = (3y)³ + (5z)³.
  2. Using a³+b³ = (a+b)(a²−ab+b²): = (3y+5z)[(3y)² − (3y)(5z) + (5z)²].
✓ 27y³ + 125z³ = (3y + 5z)(9y² − 15yz + 25z²)
40
Simplify the rational expression n³−3n²m+3nm²−m³5m²−10mn+5n², assuming the denominator is not zero.
  1. Numerator: n³ − 3n²m + 3nm² − m³ matches (n − m)³ (the (a−b)³ expansion).
  2. Denominator: 5m² − 10mn + 5n² = 5(m² − 2mn + n²) = 5(m − n)² = 5(n − m)².
  3. Expression = (n−m)³5(n−m)² = n − m5, after cancelling (n−m)².
✓ n³−3n²m+3nm²−m³5m²−10mn+5n² = n − m5
41
Without actually calculating the cubes, find the value of: −34³ + −58³ + 118³.
  1. Let a = −34, b = −58, c = 118.
  2. a + b + c = −34 − 58 + 118 = −6 − 5 + 118 = 08 = 0.
  3. Since a + b + c = 0, we have a³ + b³ + c³ = 3abc.
  4. = 3−34−58118 = 3 × 165256 = 495256.
✓ −34³ + −58³ + 118³ = 495256
42
Factor 16y² − 24y + 9 using suitable identities.
  1. Use identity (a − b)² = a² − 2ab + b².
  2. 16y² − 24y + 9 = (4y)² − 2(4y)(3) + (3)².
✓ 16y² − 24y + 9 = (4y − 3)²
43
Without finding the cubes, factorise: (2r − 3s)³ + (3s − 5t)³ + (5t − 2r)³.
  1. Let a = 2r−3s, b = 3s−5t, c = 5t−2r; then a+b+c = 2r−3s+3s−5t+5t−2r = 0.
  2. Since a+b+c = 0, a³+b³+c³ = 3abc.
  3. So (2r−3s)³+(3s−5t)³+(5t−2r)³ = 3(2r−3s)(3s−5t)(5t−2r).
✓ = 3(2r − 3s)(3s − 5t)(5t − 2r)
44
Factorize: 64a³ − 27b³ − 144a²b + 108ab²
  1. 64a³ − 27b³ − 144a²b + 108ab² = (4a)³ − (3b)³ − 3(4a)(3b)(4a − 3b).
  2. This matches (a−b)³ = a³−b³−3ab(a−b), with a → 4a, b → 3b.
✓ 64a³ − 27b³ − 144a²b + 108ab² = (4a − 3b)³
45
If a − b = 4 and ab = 45, find the value of a³ − b³.
  1. a³ − b³ = (a − b)(a² + ab + b²) = (a−b)[(a−b)² + 2ab + ab] = (a−b)[(a−b)² + 3ab].
  2. = 4 × [4² + 3(45)] = 4 × [16 + 135] = 4 × 151.
✓ a³ − b³ = 604

Section C · Factorisation / Short Answer (Q46–Q60)

3 Marks each
46
Find the value of (67.8)² − (32.2)²
  1. Using a² − b² = (a + b)(a − b), with a = 67.8, b = 32.2:
  2. (67.8)² − (32.2)² = (67.8 + 32.2)(67.8 − 32.2) = 100 × 35.6.
✓ (67.8)² − (32.2)² = 3560
47
Factorise: (2x − 3y)³ + (3y − 4z)³ + (4z − 2x)³
  1. Let a = 2x−3y, b = 3y−4z, c = 4z−2x; then a+b+c = 0.
  2. So a³+b³+c³ = 3abc.
  3. (2x−3y)³+(3y−4z)³+(4z−2x)³ = 3(2x−3y)(3y−4z)(4z−2x) = 3(2x−3y)(3y−4z)·2(2z−x).
✓ = 6(2x − 3y)(3y − 4z)(2z − x)
48
Factories by taking out common factors: ab(a² + b² − c²) − bc(c² − a² − b²) + ca(a² + b² − c²)
  1. Note that −bc(c² − a² − b²) = bc(a² + b² − c²).
  2. So the expression becomes ab(a²+b²−c²) + bc(a²+b²−c²) + ca(a²+b²−c²).
  3. Taking out the common factor (a²+b²−c²): = (a²+b²−c²)(ab + bc + ca).
✓ = (a² + b² − c²)(ab + bc + ca)
49
Factorise: x² + 14x² + 1 − 7x − 72x
  1. Write x² + 14x² + 1 = x² + [12x]² + 2·x·12x = [x + 12x]².
  2. And −7x − 72x = −7[x + 12x].
  3. So the expression = [x + 12x]² − 7[x + 12x] = [x + 12x]·[x + 12x − 7].
✓ = [x + 12x][x − 7 + 12x]
50
Factorise the following:
i. 4x³ − 6x²
ii. 21py² − 56py
  1. (i) H.C.F. of 4x³ and 6x² is 2x². So 4x³ − 6x² = 2x²(2x − 3).
  2. (ii) H.C.F. of 21py² and 56py is 7py. So 21py² − 56py = 7py(3y − 8).
✓ (i) 2x²(2x − 3); (ii) 7py(3y − 8)
51
Factorise: a² − b² − (a + b)²
  1. a² − b² − (a+b)² = a² − b² − (a² + 2ab + b²) = a² − b² − a² − 2ab − b².
  2. = −2ab − 2b² = −2b(a + b).
✓ a² − b² − (a + b)² = −2b(a + b)
52
Factorise: 4x⁴ − x² − 12x − 36
  1. Rewrite: 4x⁴ − (x² + 12x + 36) = 4x⁴ − (x + 6)² = (2x²)² − (x + 6)².
  2. Using a²−b² = (a+b)(a−b): = [2x² + (x+6)][2x² − (x+6)] = (2x²+x+6)(2x²−x−6).
  3. Factor the quadratic 2x²−x−6 = 2x²−4x+3x−6 = 2x(x−2)+3(x−2) = (x−2)(2x+3).
✓ 4x⁴ − x² − 12x − 36 = (2x² + x + 6)(x − 2)(2x + 3)
53
Factorize the following:
i. a − 8ab³
ii. x⁶ − 1
  1. (i) a − 8ab³ = a[1 − 8b³] = a[(1)³ − (2b)³].
  2. Using a³−b³ = (a−b)(a²+ab+b²): = a(1−2b)(1 + 2b + 4b²).
  3. (ii) x⁶ − 1 = (x³)² − (1)² = (x³−1)(x³+1).
  4. Further factor each: x³−1 = (x−1)(x²+x+1) and x³+1 = (x+1)(x²−x+1).
✓ (i) a(1−2b)(1+2b+4b²); (ii) (x−1)(x+1)(x²+x+1)(x²−x+1)
54
Factorise the following:
i. 4(a − 1)² − 4(a − 1) − 3
ii. 1 − 2a − 2b − 3(a + b)²
  1. (i) Let t = (a−1). Expression = 4t² − 4t − 3 = 4t² − 6t + 2t − 3 = 2t(2t−3) + 1(2t−3) = (2t−3)(2t+1).
  2. Substitute back: = [2(a−1)−3][2(a−1)+1] = (2a−5)(2a−1).
  3. (ii) Rewrite as 1 − 2(a+b) − 3(a+b)². Let t = (a+b): = 1 − 2t − 3t² = 1 − 3t + t − 3t² = (1−3t) + t(1−3t) = (1+t)(1−3t).
  4. Substitute back: = (1 + a + b)[1 − 3(a+b)] = (1 + a + b)(1 − 3a − 3b).
✓ (i) (2a − 5)(2a − 1); (ii) (1 + a + b)(1 − 3a − 3b)
55
Try to evaluate the following using a suitable identity: i. 35²   ii. 65²   iii. 85²   iv. 105². Do you observe any interesting pattern?
  1. Use (a + b)² = a² + 2ab + b².
  2. i. 35² = (30+5)² = 900 + 300 + 25 = 1225.
  3. ii. 65² = (60+5)² = 3600 + 600 + 25 = 4225.
  4. iii. 85² = (80+5)² = 6400 + 800 + 25 = 7225.
  5. iv. 105² = (100+5)² = 10000 + 1000 + 25 = 11025.
  6. Pattern: every number here ends in 5, and its square always ends in 25 — the digits before the 25 come from multiplying the number before the 5 with the next whole number (e.g. 3×4=12 for 35², 6×7=42 for 65²).
✓ 35²=1225, 65²=4225, 85²=7225, 105²=11025
56
Factorise: x² + [a²+1a]x + 1
  1. Split the middle term: x² + ax + 1ax + 1 = x² + ax + xa + 1.
  2. Group: (x² + ax) + (xa + 1) = x(x + a) + 1a(x + a).
  3. Factor out (x + a): = (x + a)(x + 1a).
✓ x² + [a²+1a]x + 1 = (x + a)(x + 1a)
57
Factorise the following:
i. a⁴ − b⁴ + 2b² − 1
ii. x³ − 25x
  1. (i) Rewrite as a⁴ − (b⁴ − 2b² + 1) = a⁴ − (b² − 1)² = (a²)² − (b²−1)².
  2. Using a²−b²=(a−b)(a+b): = (a² + b² − 1)(a² − b² + 1).
  3. (ii) x³ − 25x = x(x² − 25) = x(x² − 5²) = x(x − 5)(x + 5).
✓ (i) (a²+b²−1)(a²−b²+1); (ii) x(x−5)(x+5)
58
Factorise the following:
i. 2x² + 13xy − 24y²
ii. 6x² − 5xy − 6y²
  1. (i) Need two numbers with sum 13 and product 2×(−24)=−48: these are 16 and −3.
  2. 2x²+13xy−24y² = 2x²+16xy−3xy−24y² = 2x(x+8y) − 3y(x+8y) = (x+8y)(2x−3y).
  3. (ii) Need two numbers with sum −5 and product 6×(−6)=−36: these are −9 and 4.
  4. 6x²−5xy−6y² = 6x²−9xy+4xy−6y² = 3x(2x−3y) + 2y(2x−3y) = (2x−3y)(3x+2y).
✓ (i) (x+8y)(2x−3y); (ii) (2x−3y)(3x+2y)
59
Factorise: x⁶ + 8y⁶ − z⁶ + 6x²y²z²
  1. Write as (x²)³ + (2y²)³ + (−z²)³ − 3(x²)(2y²)(−z²).
  2. Using a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca), with a=x², b=2y², c=−z²:
  3. = (x²+2y²−z²)[(x²)²+(2y²)²+(z²)² − x²(2y²) − 2y²(−z²) − x²(−z²)]
  4. = (x²+2y²−z²)(x⁴+4y⁴+z⁴−2x²y²+2y²z²+x²z²).
✓ = (x²+2y²−z²)(x⁴+4y⁴+z⁴−2x²y²+2y²z²+x²z²)
60
Factorise the following:
i. 2x⁴ − 32
ii. a²(b + c) − (b + c)³
  1. (i) 2x⁴−32 = 2(x⁴−16) = 2[(x²)²−4²] = 2(x²−4)(x²+4) = 2(x−2)(x+2)(x²+4).
  2. (ii) a²(b+c) − (b+c)³ = (b+c)[a² − (b+c)²] = (b+c)(a+b+c)(a−b−c), using a²−b²=(a−b)(a+b).
✓ (i) 2(x−2)(x+2)(x²+4); (ii) (b+c)(a+b+c)(a−b−c)

Section D · Case Study Based (Q61–Q65)

4 Marks each
61
Case Study: A municipal corporation is redesigning a square-shaped city park that currently measures 100 m × 100 m. To improve functionality, they decide to expand the park by adding 5 m to both the length and the width, making it a larger square (105 m × 105 m). Later, to build a footpath around it, they take 2 m off the length and 2 m off the width of this expanded park, making it (105−2) m × (105−2) m.
  • Which algebraic identity is most useful to calculate the area of the initially expanded park (105 m × 105 m)? (1)
  • What is the exact area of the expanded park (105 m × 105 m) using identities? (1)
  • If the park dimensions become (105−2) × (105−2), find its area using a suitable identity. (2)
    OR — Find the total area of the footpath built around the park (the difference between the expanded park area and the final park area). (2)
  1. (i) Since 105 = 100 + 5, the side length has the form (a + b) with a = 100, b = 5. The most useful identity is (a + b)² = a² + 2ab + b².
  2. (ii) (100 + 5)² = 100² + 2(100)(5) + 5² = 10000 + 1000 + 25 = 11025 m².
  3. (iii) New side = 105 − 2 = 103 m, so area = (105 − 2)². Use (a − b)² = a² − 2ab + b² with a = 105, b = 2:
  4. (105 − 2)² = 105² − 2(105)(2) + 2² = 11025 − 420 + 4 = 10609 m².
  5. OR: Area of footpath = 11025 − 10609 = 416 m².
✓ Identity: (a±b)²; Expanded park area = 11025 m²; Final park area = 10609 m² (footpath area = 416 m²)
62
Case Study: A carpenter is building a cube-shaped wooden gift box. The volume of the box, in cubic centimetres, works out to 8y³ + 36y² + 54y + 27, where y is a design parameter he uses across different box sizes.
  • Express this volume in the form (a + b)³ by identifying a and b. (1)
  • Write down the side length of the cube in terms of y. (1)
  • If y = 3 cm, find the actual side length and the volume of the box. (2)
    OR — If the volume were instead 8y³ − 36y² + 54y − 27, what would the side length be, and why does the sign pattern in the expression change? (2)
  1. (i) Comparing the first and last terms: 8y³ = (2y)³ so a = 2y; 27 = 3³ so b = 3.
  2. Check middle terms using (a+b)³ = a³+3a²b+3ab²+b³: 3a²b = 3(2y)²(3) = 36y² ✓, 3ab² = 3(2y)(3)² = 54y ✓.
  3. So the volume = (2y + 3)³.
  4. (ii) Since volume = (side)³, side length = (2y + 3) cm.
  5. (iii) At y = 3: side = 2(3)+3 = 9 cm. Volume = 9³ = 729 cm³.
  6. OR: The expression 8y³−36y²+54y−27 matches (a−b)³ = a³−3a²b+3ab²−b³, so the side length would be (2y − 3) cm. The signs alternate ( +, −, +, − ) because substituting −b into the odd-power terms of the expansion keeps flipping their sign.
✓ Volume = (2y+3)³; Side = (2y+3) cm; at y=3: side = 9 cm, volume = 729 cm³
63
Case Study: National Association For The Blind (NAB) aims to empower and well-inform the visually challenged population of our country, enabling them to lead a life of dignity and productivity. Ravi donated ₹(x³ + 1x³) to NAB. When his cousin asked him the amount donated, he just gave the hint: x + 1x = 10.
  • When (x + a)(x + b) is expanded, what is the resulting expression in the form (x² + ______ x + ab)? (1)
  • What is the mathematical formula for the identity represented as (x − y)³? (1)
  • What is the amount donated by Ravi? (2)
    OR — What is the amount donated by Ravi if x + 1x = 7? (2)
  1. (i) (x + a)(x + b) = x² + (a + b)x + ab — the blank is filled by (a + b).
  2. (ii) (x − y)³ = x³ − y³ − 3xy(x − y) (equivalently x³ − 3x²y + 3xy² − y³).
  3. (iii) Cubing x + 1x = 10 using (a+b)³ = a³+b³+3ab(a+b): x³ + 1x³ + 3·x·1x·(x + 1x) = 1000.
  4. (x³ + 1x³) + 3(10) = 1000 ⇒ x³ + 1x³ = 1000 − 30 = 970. So Ravi donated ₹970.
  5. OR: With x + 1x = 7: (x³ + 1x³) + 3(7) = 343 ⇒ x³ + 1x³ = 343 − 21 = 322. So Ravi would have donated ₹322.
✓ (i) (a+b); (ii) (x−y)³=x³−y³−3xy(x−y); (iii) ₹970 (or ₹322 if x + 1x = 7)
64
Case Study: A housing society's square pool deck currently measures 50 m × 50 m. To create more lounging space, the committee decides to add 6 m to both the length and the width, making it a larger square (56 m × 56 m). Later, to lay a tiled walking border around this enlarged deck, they take 4 m off the length and 4 m off the width.
  • Which identity is most useful to calculate the area of the enlarged deck (56 m × 56 m)? (1)
  • Find the exact area of the enlarged deck using that identity. (1)
  • After the border is cut in, the deck becomes (56−4) m on each side. Find its new area using a suitable identity. (2)
    OR — Find the difference between the area of the enlarged deck and the final deck. (2)
  1. (i) Since 56 = 50 + 6, the area is (50 + 6)²; the useful identity is (a + b)² = a² + 2ab + b².
  2. (ii) (50+6)² = 50² + 2(50)(6) + 6² = 2500 + 600 + 36 = 3136 m².
  3. (iii) New side = 56 − 4 = 52 m, area = (56−4)². Use (a−b)² = a²−2ab+b² with a=56, b=4:
  4. (56−4)² = 56² − 2(56)(4) + 4² = 3136 − 448 + 16 = 2704 m².
  5. OR: Difference = 3136 − 2704 = 432 m² (also checks out directly: 52 × 52 = 2704 m²).
✓ Enlarged deck area = 3136 m²; Final deck area = 2704 m² (difference = 432 m²)
65
Case Study: A school garden committee decided to redesign a square flower bed of side x metres by adding a border of uniform width 3 metres on two adjacent sides. This forms a larger square plot of side (x + 3) metres, and its area can be expressed using the identity (a + b)² = a² + 2ab + b². The original square had side 7 metres.
  • Using the identity (a + b)² = a² + 2ab + b², expand (x + 3)² and find the total area of the new plot when x = 7. (1)
  • If instead the border width is b metres and x = 10, what is the value of 2xb when the total area of the new plot equals 196 m²? (Use the expansion of (x + b)².) (1)
  • A rectangular section of the plot is separated for roses. Its area is given by x² + 10x + 25 m². Factorise this expression completely and state the side length of this square rose section. (2)
    OR — Another section has area 4x² + 12x + 9 m². Factorise completely and find the perimeter of this square section in terms of x. (2)
  1. (i) (x+3)² = x² + 2(x)(3) + 3² = x² + 6x + 9. At x=7: = 49 + 42 + 9 = 100 m².
  2. (ii) (10+b)² = 100 + 20b + b² = 196 ⇒ b² + 20b − 96 = 0 ⇒ (b+24)(b−4) = 0 ⇒ b = 4 (taking the positive root).
  3. So 2xb = 2(10)(4) = 80 m².
  4. (iii) x² + 10x + 25 = x² + 2(x)(5) + 5² = (x + 5)². So the rose section is a square of side (x + 5) metres.
  5. OR: 4x²+12x+9 = (2x)²+2(2x)(3)+3² = (2x+3)². Side = (2x+3) m, so perimeter = 4(2x+3) = (8x+12) metres.
✓ (i) 100 m²; (ii) 2xb = 80 m²; (iii) side = (x+5) m (or perimeter = (8x+12) m)

Section E · Long Answer / Factorisation (Q66–Q75)

5 Marks each
66
Factorize: (x2 + y + z3)³ + (x3 − 2y3 + z)³ + (−5x6 − y3 − 4z3)³
  1. Let a = x2 + y + z3, b = x3 − 2y3 + z, c = −5x6 − y3 − 4z3.
  2. Add the x-terms: x2 + x3 − 5x6 = 3x6 + 2x6 − 5x6 = 0. Add the y-terms: y − 2y3 − y3 = 0. Add the z-terms: z3 + z − 4z3 = 0.
  3. So a + b + c = 0, and hence a³ + b³ + c³ = 3abc.
  4. = 3(x2 + y + z3)(x3 − 2y3 + z)(−5x6 − y3 − 4z3).
✓ = 3(x2 + y + z3)(x3 − 2y3 + z)(−5x6 − y3 − 4z3)
67
Factorise: 14(a + b)² − 916(2a − b)²
  1. Factor out 14: = 14[(a+b)² − 94(2a−b)²] = 14[(a+b)² − {32(2a−b)}²].
  2. Using a²−b²=(a+b)(a−b): = 14[(a+b) + 32(2a−b)]·[(a+b) − 32(2a−b)].
  3. First bracket: (a+b) + 3a − 32b = 4a − b2 = 8a − b2.
  4. Second bracket: (a+b) − 3a + 32b = −2a + 5b2 = 5b − 4a2.
  5. So the product = 14 × 8a−b2 × 5b−4a2 = (8a−b)(5b−4a)16.
✓ 14(a+b)² − 916(2a−b)² = 116(8a − b)(5b − 4a)
68
Prove that for any non-zero variables x and y, the expression (x+y)³ − (x−y)³2y(3x²+y²) simplifies to 1.
  1. Expand: (x+y)³ = x³+3x²y+3xy²+y³, and (x−y)³ = x³−3x²y+3xy²−y³.
  2. Numerator = (x+y)³ − (x−y)³ = (x³+3x²y+3xy²+y³) − (x³−3x²y+3xy²−y³).
  3. Grouping like terms: the x³ and 3xy² terms cancel; we're left with 6x²y + 2y³.
  4. Numerator = 6x²y + 2y³ = 2y(3x² + y²).
  5. So the expression = 2y(3x²+y²)2y(3x²+y²). Since x, y ≠ 0, this common non-zero factor cancels completely, leaving 1.
✓ The expression simplifies to 1 — proved.
69
Factorise: a²b + ab² − abc − b²c + axy + bxy
  1. Group the six terms into two sets of three: (a²b − abc + axy) + (ab² − b²c + bxy).
  2. Factor each group: a(ab − bc + xy) + b(ab − bc + xy).
  3. Factor out the common bracket (ab − bc + xy): = (a + b)(ab − bc + xy).
✓ a²b+ab²−abc−b²c+axy+bxy = (a + b)(ab − bc + xy)
70
Factorise the following:
  • (a + b)³ − a − b
  • x² − 2xy + y² − a² − 2ab − b²
  1. (i) (a+b)³ − a − b = (a+b)³ − (a+b). Take out the common factor (a+b): = (a+b)[(a+b)² − 1].
  2. Using a²−b²=(a−b)(a+b) on [(a+b)²−1²]: = (a+b)(a+b−1)(a+b+1).
  3. (ii) Rewrite −a²−2ab−b² as −(a²+2ab+b²) = −(a+b)², so the expression is (x²−2xy+y²) − (a+b)² = (x−y)² − (a+b)².
  4. Using a²−b²=(a−b)(a+b): = (x−y−a−b)(x−y+a+b).
✓ (i) (a+b)(a+b−1)(a+b+1); (ii) (x−y−a−b)(x−y+a+b)
71
Factorise the following:
  • x³ + x² − 1x² + 1x³
  • (x + 1)⁶ − (x − 1)⁶
  1. (i) Group as (x³ + 1x³) + (x² − 1x²).
  2. Using a³+b³=(a+b)(a²−ab+b²) and a²−b²=(a+b)(a−b), both with a = x, b = 1x, both groups share the factor (x + 1x):
  3. = (x + 1x)(x² − 1 + 1x²) + (x + 1x)(x − 1x) = (x + 1x)[x² + 1x² − 1 + x − 1x].
  4. (ii) (x+1)⁶ − (x−1)⁶ = [(x+1)³]² − [(x−1)³]² = [(x+1)³+(x−1)³]·[(x+1)³−(x−1)³].
  5. Using a³+b³=(a+b)(a²−ab+b²) with a=x+1,b=x−1: sum = 2x·[(x+1)²−(x+1)(x−1)+(x−1)²] = 2x(x²+3).
  6. Using a³−b³=(a−b)(a²+ab+b²): difference = 2·[(x+1)²+(x+1)(x−1)+(x−1)²] = 2(3x²+1).
  7. Product = 2x(x²+3) × 2(3x²+1) = 4x(x²+3)(3x²+1).
✓ (i) (x + 1x)[x² + 1x² − 1 + x − 1x]; (ii) 4x(x²+3)(3x²+1)
72
Factorize the following:
  • (a² − b²)(c² − d²) − 4abcd
  • 4x² − 12ax − y² − z² − 2yz + 9a²
  1. (i) Expand: a²c² − a²d² − b²c² + b²d² − 4abcd.
  2. Regroup: (a²c² + b²d² − 2abcd) − (a²d² + b²c² + 2abcd) = (ac − bd)² − (ad + bc)².
  3. Using a²−b²=(a+b)(a−b): = (ac−bd+ad+bc)(ac−bd−ad−bc).
  4. (ii) Rearrange: 9a² + 4x² − 12ax − (y²+z²+2yz) = (3a)² + (2x)² − 2(3a)(2x) − (y+z)² = (2x − 3a)² − (y+z)².
  5. Using a²−b²=(a+b)(a−b): = (2x − 3a + y + z)(2x − 3a − y − z).
✓ (i) (ac−bd+ad+bc)(ac−bd−ad−bc); (ii) (2x−3a+y+z)(2x−3a−y−z)
73
Factorise: a⁹ − 1
  1. a⁹ − 1 = (a³)³ − 1³.
  2. Using a³−b³=(a−b)(a²+ab+b²) with a→a³, b→1: = (a³−1)[(a³)²+a³+1] = (a³−1)(a⁶+a³+1).
  3. Further factor (a³−1) = (a−1)(a²+a+1).
✓ a⁹ − 1 = (a − 1)(a² + a + 1)(a⁶ + a³ + 1)
74
Factorise: a³ + 2a² + 5a + 10
  1. Group in pairs: (a³ + 2a²) + (5a + 10).
  2. Factor each group: a²(a + 2) + 5(a + 2).
  3. Factor out the common bracket (a + 2): = (a² + 5)(a + 2).
✓ a³ + 2a² + 5a + 10 = (a² + 5)(a + 2)
75
If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25
  1. Use the identity a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca) = (a+b+c)[a²+b²+c² − (ab+bc+ca)].
  2. = 5[a²+b²+c² − 10] (substituting a+b+c=5 and ab+bc+ca=10).
  3. Now square (a+b+c)=5: (a+b+c)² = a²+b²+c²+2(ab+bc+ca) ⇒ 25 = a²+b²+c²+2(10).
  4. a²+b²+c² = 25 − 20 = 5.
  5. So a³+b³+c³−3abc = 5(5 − 10) = 5(−5) = −25.
✓ a³ + b³ + c³ − 3abc = −25 — proved.
Prepared by Sumeet Sahu · Mob: 8103405051 · Unique Study Point
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📋 Details

ClassClass IX (CBSE / NCERT)
SubjectMaths
ChapterChapter 4: Exploring Algebraic Identities
Resource TypeWorksheet
Last Updated05 September 2026
Session2026-27 (Latest NCERT Syllabus)
Downloads0+
Prepared bySumeet Sahu, Unique Study Point, Indore
CostFree
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