Class 9 Maths Ganita Manjari Chapter 4 Solutions (NCERT 2026–27) – Exploring Algebraic Identities

These Class 9 Maths Ganita Manjari Chapter 4 solutions cover Exploring Algebraic Identities from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step so you can understand each identity and revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 4 Exercises: 4.1–4.5, End-of-Chapter Session: 2026–27
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Chapter 4 Overview

Chapter 4 of Ganita Manjari, Exploring Algebraic Identities, develops the standard algebraic identities and shows three uses for each: fast numerical calculation, expanding expressions, and factorising them (including with algebra tiles). It then extends to the cubic identities and to simplifying rational expressions by cancelling common factors. The Class 9 Maths Ganita Manjari Chapter 4 solutions below work through every exercise step by step.

Key Concepts & Definitions

Identity: an equation that is true for all values of the variables (e.g. (a + b)2 = a2 + 2ab + b2).

Expanding: using an identity to remove brackets.

Factorising: writing an expression as a product of factors, often by recognising an identity or by splitting the middle term.

Splitting the middle term: for x2 + (a + b)x + ab, find a and b with sum = coefficient of x and product = constant.

Rational expression: a ratio of two polynomials; simplify it by factorising and cancelling common factors (denominator ≠ 0).

All Identities Used in This Chapter

(a + b)2 = a2 + 2ab + b2  •  (a − b)2 = a2 − 2ab + b2

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

a2 − b2 = (a + b)(a − b)  •  (x + a)(x + b) = x2 + (a + b)x + ab

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

x3 − y3 = (x − y)(x2 + xy + y2)  •  x3 + y3 = (x + y)(x2 − xy + y2)

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

Exercise Set 4.1

1. Using the identity (a + b)2 = a2 + 2ab + b2, expand the following: (i) (7x + 4y)2   (ii) ((7/5)x + (3/2)y)2   (iii) (2.5p + 1.5q)2 (iv) ((3/4)s + 8t)2   (v) (x + 1/(2y))2   (vi) (1/x + 1/y)2

SOLUTION (i) 49x2 + 56xy + 16y2. (ii) (49/25)x2 + (21/5)xy + (9/4)y2. (iii) 6.25p2 + 7.5pq + 2.25q2. (iv) (9/16)s2 + 12st + 64t2. (v) x2 + x/y + 1/(4y2). (vi) 1/x2 + 2/(xy) + 1/y2.

2. Using the same identity, find the values of the following: (i) (64)2   (ii) (105)2   (iii) (205)2

SOLUTION (i) (60 + 4)2 = 3600 + 480 + 16 = 4096. (ii) (100 + 5)2 = 10000 + 1000 + 25 = 11025. (iii) (200 + 5)2 = 40000 + 2000 + 25 = 42025.
Think and Reflect What if we replace b by −b in (a + b)2 = a2 + 2ab + b2? Answer. We get (a − b)2 = a2 − 2ab + b2 — another identity, used the same way.

Exercise Set 4.2

1. Factor completely: (i) 9x2 + 24xy + 16y2   (ii) 4s2 + 20st + 25t2   (iii) 49x2 + 28xy + 4y2 (iv) 64p2 + (32/3)pq + (4/9)q2   *(v) 3a2 + 4ab + (4/3)b2   *(vi) (9/5)s2 + 6sv + 5v2

SOLUTION (i) (3x + 4y)2. (ii) (2s + 5t)2. (iii) (7x + 2y)2. (iv) (8p + (2/3)q)2  [= (4/9)(12p + q)2]. *(v) Take out 1/3: 3a2 + 4ab + (4/3)b2 = (1/3)(9a2 + 12ab + 4b2) = (1/3)(3a + 2b)2. *(vi) Take out 1/5: (9/5)s2 + 6sv + 5v2 = (1/5)(9s2 + 30sv + 25v2) = (1/5)(3s + 5v)2.

2. Find the values using the identity (a − b)2 = a2 − 2ab + b2: (i) (79)2   (ii) (193)2   (iii) (299)2

SOLUTION (i) (80 − 1)2 = 6400 − 160 + 1 = 6241. (ii) (200 − 7)2 = 40000 − 2800 + 49 = 37249. (iii) (300 − 1)2 = 90000 − 600 + 1 = 89401.

Exercise Set 4.3

1. Find the following squares using one of the identities (state which is easier): (i) 1172   (ii) 782   (iii) 1982   (iv) 2142   (v) 11042   (vi) 11202

SOLUTION (i) (120 − 3)2 = 14400 − 720 + 9 = 13689. (ii) (80 − 2)2 = 6400 − 320 + 4 = 6084. (iii) (200 − 2)2 = 40000 − 800 + 4 = 39204. (iv) (200 + 14)2 = 40000 + 5600 + 196 = 45796. (v) (1100 + 4)2 = 1210000 + 8800 + 16 = 1218816. (vi) (1100 + 20)2 = 1210000 + 44000 + 400 = 1254400.

2. Factor using suitable identities: (i) 16y2 − 24y + 9   (ii) (9/4)s2 + 6st + 4t2   (iii) m2/9 + mk/3 + k2/4 + 3nk + 2mn + 9n2 (iv) p2/16 − 2 + 16/p2   (v) 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc

SOLUTION (i) (4y − 3)2. (ii) ((3/2)s + 2t)2. (iii) (m/3 + k/2 + 3n)2. (iv) (p/4 − 4/p)2. (v) (3a − 2b + c)2.

3. Expand using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)2   (ii) (3x − 2y + 4z)2

SOLUTION (i) p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr. (ii) 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24xz.

4. Is this an identity? (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = 2a2 + 2b2 + 2c2.

SOLUTION Expanding the left side gives 3a2 + 3b2 + 3c2 − 2ab − 2bc − 2ca, which is not equal to 2a2 + 2b2 + 2c2. So No, it is not an identity (it holds only for special values, not all a, b, c).
Think and Reflect Evaluate 352, 652, 852, 1052. Do you observe a pattern? Answer. 352 = 1225, 652 = 4225, 852 = 7225, 1052 = 11025. For a number ending in 5, say “n5”, the square is n(n + 1) followed by 25 — e.g. 852: 8×9 = 72, then 25 → 7225.

Exercise Set 4.4

1. Fill in the blanks to complete the following identities: (i) s2 − 11s + 24 = (____)(____)   (ii) (____)(x + 1) = (3x2 − 4x − 7) (iii) 10x2 − 11x − 6 = (2x − ___)(___ + 2)   (iv) 6x2 + 7x + 2 = (____)(____)

SOLUTION (i) (s − 3)(s − 8). (ii) (3x − 7)(x + 1). (iii) (2x − 3)(5x + 2). (iv) (2x + 1)(3x + 2).

2. Use a suitable identity to find these products without multiplying directly: (i) (41)2   (ii) (27)2   (iii) 23 × 17   (iv) (135)2   (v) (97)2   (vi) 18 × 29   (vii) 34 × 43   (viii) (205)2

SOLUTION (i) (40 + 1)2 = 1681. (ii) (30 − 3)2 = 729. (iii) (20 + 3)(20 − 3) = 400 − 9 = 391. (iv) (130 + 5)2 = 18225. (v) (100 − 3)2 = 9409. (vi) 18 × 29 = (23.5)2 − (5.5)2 = 552.25 − 30.25 = 522. (vii) 34 × 43 = (38.5)2 − (4.5)2 = 1462. (viii) (200 + 5)2 = 42025.

3. Factor the following: (i) 9a2 + b2 + 4c2 − 6ab + 12ac − 4bc   (ii) 16s2 + 25t2 − 40st   (iii) r2 − r − 42 (iv) 49g2 + 14gh + h2   (v) 64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw

SOLUTION (i) (3a − b + 2c)2. (ii) (4s − 5t)2. (iii) (r − 7)(r + 6). (iv) (7g + h)2. (v) (8u − 11v − 2w)2.
Think and Reflect James writes (a − b)2(a + b) = (a2 − 2ab + b2)(a + b); Reshma writes it as (a − b)[(a − b)(a + b)] = (a − b)(a2 − b2). Who is correct? Answer. Both are correct — they are two valid routes to the same product; Reshma’s grouping is just a quicker way, and both give a3 − a2b − ab2 + b3.

Exercise Set 4.5

1. Simplify the following rational expressions (denominators ≠ 0): (i) (3p2 − 3pq − 18q2) / (p2 + 3pq − 10q2)   (ii) (n3 − 3n2m + 3nm2 − m3) / (5m2 − 10mn + 5n2) (iii) (w3 − v3 + x3 + 3wvx) / (w2 + v2 + x2 − 2wv − 2vx + 2wx)   (iv) (4y2 − 20yz + 25z2) / (25z2 − 4y2) (v) ((x2 + x − 6)(x2 − 7x + 12)) / ((x2 − 6x + 8)(x2 − 9))   (vi) (p4 − 16) / (p2 − 4p + 4)

SOLUTION (i) Factor: numerator = 3(p − 3q)(p + 2q), denominator = (p + 5q)(p − 2q). They share no common factor, so the expression is already in lowest terms: 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]. (ii) Numerator = (n − m)3, denominator = 5(n − m)2 ⇒ (n − m)/5. (iii) Numerator = (w − v + x)(w2 + v2 + x2 + wv + vx − wx), denominator = (w − v + x)2 ⇒ (w2 + v2 + x2 + wv + vx − wx) / (w − v + x). (iv) Numerator = (5z − 2y)2, denominator = (5z − 2y)(5z + 2y) ⇒ (5z − 2y)/(5z + 2y). (v) = [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)] = 1. (vi) p4 − 16 = (p − 2)(p + 2)(p2 + 4), denominator = (p − 2)2 ⇒ (p + 2)(p2 + 4)/(p − 2).

Class 9 Maths Ganita Manjari Chapter 4 Solutions — End-of-Chapter Exercises

1. Use suitable identities to find the following products: (i) (–3x + 4)2   (ii) (2s + 7)(2s − 7)   (iii) (p2 + 1/2)(p2 − 1/2)   (iv) (2n + 7)(2n − 7) (v) (s − 2t)(s2 + 2st + 4t2)   (vi) (1/(2r) − 4r)2   (vii) (–3m + 4k − l)2   (viii) (x − (1/3)y)3   (ix) ((7/2)k − (2/3)m)3

SOLUTION (i) 9x2 − 24x + 16. (ii) 4s2 − 49. (iii) p4 − 1/4. (iv) 4n2 − 49. (v) s3 − 8t3  [(a − b)(a2 + ab + b2) with a = s, b = 2t]. (vi) 16r2 − 4 + 1/(4r2). (vii) 9m2 + 16k2 + l2 − 24mk − 8kl + 6lm. (viii) x3 − x2y + (1/3)xy2 − (1/27)y3. (ix) (343/8)k3 − (49/2)k2m + (14/3)km2 − (8/27)m3.

2. Find the values using suitable identities: (i) 17 × 21   (ii) 104 × 96   (iii) 24 × 16   (iv) 1473   (v) 1993   (vi) 1273   (vii) (–107)3   (viii) (–299)3

SOLUTION (i) (19 − 2)(19 + 2) = 361 − 4 = 357. (ii) (100 + 4)(100 − 4) = 9984. (iii) (20 + 4)(20 − 4) = 384. (iv) (150 − 3)3 = 3176523. (v) (200 − 1)3 = 7880599. (vi) (130 − 3)3 = 2048383. (vii) −(1073) = −1225043. (viii) −(2993) = −26730899.

3. Factor the following algebraic expressions: (i) 4y2 + 1 + 1/(16y2)   (ii) 9m2 − 1/(25n2)   (iii) 27b3 − 1/(64b3)   (iv) x2 + 5x/6 + 1/6 (v) 27u3 − 1/125 − 27u2/5 + 9u/25   (vi) 64y3 + (1/125)z3   (vii) p3 + 27q3 + r3 − 9pqr   (viii) 9m2 − 12m + 4 (ix) 9x3 − (8/3)y3 + (1/3)z3 + 6xyz   (x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy   (xi) 27u3 − 1/216 − 9u2/2 + u/4

SOLUTION (i) (2y + 1/(4y))2. (ii) (3m − 1/(5n))(3m + 1/(5n)). (iii) (3b − 1/(4b))(9b2 + 3/4 + 1/(16b2)). (iv) (x + 1/2)(x + 1/3). (v) (3u − 1/5)3. (vi) (4y + z/5)(16y2 − 4yz/5 + z2/25). (vii) (p + 3q + r)(p2 + 9q2 + r2 − 3pq − 3qr − pr). (viii) (3m − 2)2. (ix) Multiply through by 3: 27x3 − 8y3 + z3 + 18xyz = (3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3xz). So the expression = (1/3)(3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3xz). (x) This is meant to be a perfect-square trinomial; with the cross-terms 12xy + 24xz + 36yz it is (2x + 3y + 6z)2. (As printed, two cross-term coefficients appear transposed; the intended factorisation is (2x + 3y + 6z)2.) (xi) (3u − 1/6)3.

4. Simplify the following (denominators ≠ 0): (i) (4x2 + 4x + 1) / (4x2 − 1)   (ii) 9(3a3 − 24b3) / (9a2 − 36b2)   (iii) (s3 + 125t3) / (s2 − 2st − 35t2)

SOLUTION (i) (2x + 1)2 / [(2x − 1)(2x + 1)] = (2x + 1)/(2x − 1). (ii) 27(a − 2b)(a2 + 2ab + 4b2) / [9(a − 2b)(a + 2b)] = 3(a2 + 2ab + 4b2)/(a + 2b). (iii) (s + 5t)(s2 − 5st + 25t2) / [(s − 7t)(s + 5t)] = (s2 − 5st + 25t2)/(s − 7t).

5. Find possible expressions for the length and breadth of rectangles with these areas: (i) 25a2 − 30ab + 9b2   (ii) 36s2 − 49t2

SOLUTION (i) 25a2 − 30ab + 9b2 = (5a − 3b)2 ⇒ length = breadth = (5a − 3b). (ii) 36s2 − 49t2 = (6s + 7t)(6s − 7t) ⇒ length = (6s + 7t), breadth = (6s − 7t).

6. Find possible expressions for the length, breadth and height of cuboids with these volumes: (i) 6a2 − 24b2   (ii) 3ps2 − 15ps + 12p

SOLUTION (i) 6a2 − 24b2 = 6(a − 2b)(a + 2b) ⇒ dimensions 6, (a − 2b), (a + 2b). (ii) 3ps2 − 15ps + 12p = 3p(s − 1)(s − 4) ⇒ dimensions 3p, (s − 1), (s − 4).

7. The village playground is a square of side 40 m. A path of width s metres is created around it. Find an expression for the area of the path in terms of s.

SOLUTION Outer square side = (40 + 2s). Path area = (40 + 2s)2 − 402 = 1600 + 160s + 4s2 − 1600 = 4s2 + 160s (= 4s(s + 40)) sq. m.

8. If a number plus its reciprocal equals 10/3, find the number.

SOLUTION x + 1/x = 10/3 ⇒ 3x2 − 10x + 3 = 0 ⇒ (3x − 1)(x − 3) = 0 ⇒ x = 3 or 1/3.

9. A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length.

SOLUTION 2x2 + 7x + 3 = (2x + 1)(x + 3). Dividing by the width (2x + 1) gives length = (x + 3) hastas.

*10. If both x − 2 and x − 1/2 are factors of px2 + 5x + r, show that p = r.

SOLUTION x = 2 is a root: 4p + 10 + r = 0 ⇒ 4p + r = −10. x = 1/2 is a root: p/4 + 5/2 + r = 0 ⇒ p + 4r = −10. Subtracting the two equations: 3p − 3r = 0 ⇒ p = r.

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

SOLUTION a3 + b3 + c3 − 3abc = (a + b + c)[(a + b + c)2 − 3(ab + bc + ca)] = 5[25 − 30] = 5(−5) = −25. ✓

*12. By factoring, check that n3 − n is always divisible by 6 for all natural numbers n.

SOLUTION n3 − n = n(n2 − 1) = (n − 1)·n·(n + 1) — the product of three consecutive integers. Among any three consecutive integers, one is divisible by 3 and at least one is even (divisible by 2). So the product is divisible by 2 × 3 = 6.

*13. Find the value of: (i) x3 + y3 − 12xy + 64, when x + y = −4   (ii) x3 − 8y3 − 36xy − 216, when x = 2y + 6

SOLUTION (i) x3 + y3 + 43 − 3·x·y·4 = (x + y + 4)(…). Since x + y = −4, x + y + 4 = 0, so the value is 0. (ii) x3 + (−2y)3 + (−6)3 − 3·x·(−2y)·(−6) = (x − 2y − 6)(…). Since x = 2y + 6, x − 2y − 6 = 0, so the value is 0.

Common Mistakes to Avoid

Watch out for these

  • Writing (a + b)2 = a2 + b2 — the middle term 2ab must be there.
  • Sign of the middle term: (a − b)2 = a2 − 2ab + b2.
  • In (a + b + c)2, forgetting the three cross terms 2ab, 2bc, 2ca (mind their signs).
  • For (a ± b)3, missing the 3a2b and 3ab2 terms (coefficients 1, 3, 3, 1).
  • When splitting the middle term, both sum and product conditions must hold together.
  • While simplifying rational expressions, cancel only common factors, never individual terms.

Practice MCQs & Assertion–Reason

1. (a + b)2 equals:

(a) a2 + b2    (b) a2 + 2ab + b2    (c) a2 − 2ab + b2    (d) 2ab

2. (a − b)2 equals:

(a) a2 − b2    (b) a2 + 2ab + b2    (c) a2 − 2ab + b2    (d) a2 − b2 + 2ab

3. a2 − b2 equals:

(a) (a − b)2    (b) (a + b)2    (c) (a + b)(a − b)    (d) (a + b)(a + b)

4. (x + 3)(x + 4) equals:

(a) x2 + 12    (b) x2 + 7x + 12    (c) x2 + 12x + 7    (d) x2 + 7x + 7

5. The factors of x2 − 9 are:

(a) (x − 3)(x − 3)    (b) (x + 3)(x + 3)    (c) (x − 3)(x + 3)    (d) (x − 9)(x + 1)

6. (a + b)3 equals:

(a) a3 + b3    (b) a3 + 3a2b + 3ab2 + b3    (c) a3 − b3    (d) a3 + 3ab + b3

7. Using an identity, 992 equals:

(a) 9801    (b) 9081    (c) 9999    (d) 9081

8. x2 + 10x + 25 factorises to:

(a) (x + 5)2    (b) (x − 5)2    (c) (x + 25)(x + 1)    (d) (x + 5)(x − 5)

9. (x − y)(x2 + xy + y2) equals:

(a) x3 + y3    (b) x3 − y3    (c) (x − y)3    (d) x2 − y2

10. If a + b = 5 and ab = 6, then a2 + b2 equals:

(a) 11    (b) 13    (c) 25    (d) 37

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(c), 6-(b), 7-(a), 8-(a), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: (a + b)2 = a2 + 2ab + b2.

Reason: This equation is an identity, true for all real values of a and b.

A-R 2. Assertion: x2 − 16 = (x − 4)(x + 4).

Reason: a2 − b2 = (a + b)(a − b).

A-R 3. Assertion: (a − b)2 = a2 + 2ab + b2.

Reason: The square of (a − b) expands to a2 − 2ab + b2.

A-R 4. Assertion: x2 + 5x + 6 = (x + 2)(x + 3).

Reason: We need two numbers with sum 5 and product 6, namely 2 and 3.

A-R 5. Assertion: If a + b + c = 0, then a3 + b3 + c3 = 3abc.

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

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • An identity is true for all values of the variables.
  • Squares: (a ± b)2 = a2 ± 2ab + b2; (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
  • Difference of squares: a2 − b2 = (a + b)(a − b).
  • Product form: (x + a)(x + b) = x2 + (a + b)x + ab — basis of “splitting the middle term”.
  • Cubes: (a ± b)3; x3 ± y3 = (x ± y)(x2 ∓ xy + y2).
  • a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca); if a + b + c = 0 then a3 + b3 + c3 = 3abc.
  • Simplify rational expressions by factorising and cancelling common factors.

How to score full marks in this chapter

Memorise the identities and always write the one you are using before substituting. For calculations like 1172, split the number as (round ± small). For factorising, first check for a common factor and a recognisable identity; otherwise split the middle term. For rational expressions, factor numerator and denominator fully before cancelling.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 4 about?

It develops the algebraic identities — squares, difference of squares, and cubes, including a3 + b3 + c3 − 3abc — and uses them to expand, factorise and simplify expressions.

What is the identity for (a + b) squared?

(a + b)2 = a2 + 2ab + b2.

What is a3 + b3 + c3 − 3abc?

It equals (a + b + c)(a2 + b2 + c2 − ab − bc − ca). If a + b + c = 0, then a3 + b3 + c3 = 3abc.

Are these Class 9 Maths Ganita Manjari Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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