NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities

Class 9 Ganita Manjari Chapter 4 Quick Links:

• Exercise Set 4.1
• Exercise Set 4.2
• Exercise Set 4.3
• Exercise Set 4.4
• Exercise Set 4.5
• End-of-Chapter Exercises

• List of Formulae

The chapter encourages students to not just memorise these identities, but to visualise them geometrically using squares, rectangles & cubes and to apply them in factorisation, simplification of rational expressions and fast numerical calculations. From the classic (a + b)² expansion to the powerful identity x³ + y³ + z³ − 3xyz, this chapter equips Grade 9 learners with a powerful algebraic toolkit for higher mathematics. Whether you are a student preparing for exams, a teacher planning lessons or a parent helping your child at home, this page provides everything you need for Chapter 4 of Ganita Manjari — clearly explained, well-structured and aligned with the latest NCERT 2026-27 syllabus.

Chapter 4, “Exploring Algebraic Identities,” introduces students to the concept of algebraic identities — equations that remain true for every possible value of the variables. The chapter covers visualisation of identities using geometric models, factorisation of algebraic expressions, discovering new identities through distributive property, simplification of rational expressions and real-life word problems involving factorisation.
The chapter is divided into the following key sections:

• Section 4.1: Introduction: Explores numerical patterns to build curiosity around algebraic identities. A famous example – taking three consecutive square numbers, adding the smallest and largest, then subtracting twice the middle – always results in 2.
• Section 4.2: Visualising Identities: Uses geometric squares and rectangles to prove (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
• Section 4.3: Factorisation Using Identities: Shows how known identities can be reversed to factor expressions like 36x² + 12x + 1 and 50p² + 60pq + 18q².
• Section 4.4: More Identities: Introduces (a + b + c)² and the sum/difference of cubes identities, along with the identity x³ + y³ + z³ − 3xyz.
• Section 4.5: Factorisation Using Algebra Tiles: Visualises factorisation of quadratic expressions like x² + 7x + 12 using physical algebra tile arrangements.
• Section 4.6: Factorisation Without Algebra Tiles: Teaches the splitting-the-middle-term method algebraically.
• Section 4.7: Finding New Identities: Derives (a + b)³, (a − b)³, and x³ − y³ = (x − y)(x² + xy + y²) from known identities.
• Section 4.8: Simplifying Rational Expressions: Applies factorisation to simplify algebraic fractions by cancelling common factors.

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.1 Solutions

(i) (7x + 4y)²
Answer:
Using (a + b)² = a² + 2ab + b²
Here, a = 7x and b = 4y
= (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²

(ii) [(7/5)x + (3/2)y]²
Answer:
Here, a = (7/5)x and b = (3/2)y
= [(7/5)x]² + 2 × (7/5)x × (3/2)y + [(3/2)y]²
= (49/25)x² + 2 × (21/10)xy + (9/4)y²
= (49/25)x² + (42/10)xy + (9/4)y²
Simplify:
= (49/25)x² + (21/5)xy + (9/4)y²

(iii) (2.5p + 1.5q)²
Answer:
Here, a = 2.5p and b = 1.5q
= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²

(iv) [(3/4)s + 8t]²
Answer:
Here, a = (3/4)s and b = 8t
= [(3/4)s]² + 2 × (3/4)s × 8t + (8t)²
= (9/16)s² + 2 × (24/4)st + 64t²
= (9/16)s² + 12st + 64t²

(v) [x + 1/(2y)]²
Answer:
Here, a = x and b = 1/(2y)
= x² + 2 × x × [1/(2y)] + [1/(2y)]²
= x² + (2x)/(2y) + 1/(4y²)
= x² + x/y + 1/(4y²)

(vi) ( 1/x + 1/y )²
Answer:
Here, a = 1/x and b = 1/y
= (1/x)² + 2(1/x)(1/y) + (1/y)²
= 1/x² + 2/(xy) + 1/y²

(i) (64)²
Answer:
Write 64 = 60 + 4

Using (a + b)² = a² + 2ab + b²:
(60 + 4)² = 60² + 2 × 60 × 4 + 4²
= 3600 + 480 + 16
= 4096

(ii) (105)²
Answer:
Writing 105 = 100 + 5
Now (105)² = (100 + 5)²
= 100² + 2 × 100 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]
= 10000 + 1000 + 25
= 11025

(iii) (205)²
Answer:
Write 205 = 200 + 5
So, (205)² = (200 + 5)²
= 200² + 2 × 200 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]
= 40000 + 2000 + 25
= 42025

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.2 Solutions

(i) 9x² + 24xy + 16y²
Answer:
We know that:
9x² = (3x)²
16y² = (4y)²
24xy = 2 × 3x × 4y
So, 9x² + 24xy + 16y²
= (3x)² + 2 × 3x × 4y + (4y)²
= (3x + 4y)² [Using a² + 2ab + b² = (a + b)²]

(ii) 4s² + 20st + 25t²
Answer:
We can write:
4s² = (2s)²
25t² = (5t)²
20st = 2 × 2s × 5t
So, 4s² + 20st + 25t²
= (2s)² + 2 × 2s × 5t + (5t)²
= (2s + 5t)² [Using a² + 2ab + b² = (a + b)²]

(iii) 49x² + 28xy + 4y²
Answer:
49x² = (7x)²
4y² = (2y)²
28xy = 2 × 7x × 2y
So, 49x² + 28xy + 4y²
= (7x)² + 2 × 7x × 2y + (2y)²
= (7x + 2y)² [Using a² + 2ab + b² = (a + b)²]

(iv) 64p² + (32/3)pq + (4/9)q²
Answer:
64p² = (8p)²
(4/9)q² = (2/3 q)²
Middle term:
2 × 8p × (2/3 q) = 32/3 pq
So, 64p² + (32/3)pq + (4/9)q²
= (8p)² + 2 × 8p × (2/3 q) + (2/3 q)²
= (8p + 2/3 q)² [Using a² + 2ab + b² = (a + b)²]

(v) 3a² + 4ab + (4/3)b²
Answer:
Take common factor 3:
= 3[a² + (4/3)ab + (4/9)b²]
Now, a² = (a)²
(4/9)b² = (2/3 b)²
(4/3)ab = 2 × a × (2/3 b)
So, 3a² + 4ab + (4/3)b²
= 3[a² + (4/3)ab + (4/9)b²]
= 3[a² + 2 × a × (2/3 b) + (2/3 b)²]
= 3(a + 2/3 b)² [Using a² + 2ab + b² = (a + b)²]

(vi) (9/5)s² + 6sv + 5v²
Answer:
Take common factor 1/5:
= (1/5)[9s² + 30sv + 25v²]
Now, 9s² = (3s)²
25v² = (5v)²
30sv = 2 × 3s × 5v
So, (9/5)s² + 6sv + 5v²
= (1/5)[9s² + 30sv + 25v²]
= (1/5)[(3s)² + 2 × 3s × 5v + (5v)²]
= (1/5)(3s + 5v)² [Using a² + 2ab + b² = (a + b)²]

(i) (79)²
Answer:
79 = 80 − 1
= (80 − 1)²
= 80² − 2×80×1 + 1² [Using (a − b)² = a² − 2ab + b²]
= 6400 − 160 + 1
= 6241

(ii) (193)²
Answer:
193 = 200 − 7
= (200 − 7)²
= 200² − 2×200×7 + 7² [Using (a − b)² = a² − 2ab + b²]
= 40000 − 2800 + 49
= 37249

(iii) (299)²
Answer:
299 = 300 − 1
= (300 − 1)²
= 300² − 2×300×1 + 1² [Using (a − b)² = a² − 2ab + b²]
= 90000 − 600 + 1
= 89401

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.3 Solutions

(i) 117²
Answer:
117 = 110 + 7
So, 117² = (110 + 7)²
= 110² + + 2(110)(7) + 7² [Using (a + b)² = a² + 2ab + b²]
= 12100 + 1540 + 49
= 13689

(ii) 78²
Answer:
78 = 80 – 2
So, 78² = (80 – 2)²
= 80² – 2(80)(2) + 2² [Using (a − b)² = a² − 2ab + b²]
= 6400 – 320 + 4
= 6084

(iii) 198²
Answer:
198 = 200 – 2
So, 198² = (200 – 2)²
= 200² – 2(200)(2) + 2² [Using (a − b)² = a² − 2ab + b²]
= 40000 – 800 + 4
= 39204

(iv) 214²
Answer:
214 = 200 + 14
So, 214² = (200 + 14)²
= 200² + 2(200)(14) + 14² [Using (a − b)² = a² − 2ab + b²]
= 40000 + 5600 + 196
= 45796

(v) 1104²
Answer:
1104 = 1100 + 4
So, 1104² = (1100 + 4)²
= 1100² + 2(1100)(4) + 4² [Using (a + b)² = a² + 2ab + b²]
= 1210000 + 8800 + 16
= 1218816

(vi) 1120²
Answer:
1120 = 1100 + 20
So, 1120² = (1100 + 20)²
= 1100² + 2(1100)(20) + 20² [Using (a + b)² = a² + 2ab + b²]
= 1210000 + 44000 + 400
= 1254400

(i) 16y² – 24y + 9
Answer:
16y² = (4y)²
9 = 3²
-24y = -2(4y)(3)
Therefore, 16y² – 24y + 9
= (4y)² -2(4y)(3) + 3²
= (4y – 3)² [Using a² – 2ab + b² = (a – b)²]

(ii) (9/4)s² + 6st + 4t²
Answer:
(9/4)s² = [(3s)/2]²
4t² = (2t)²
6st = 2 × (3s/2) × (2t)
Therefore, (9/4)s² + 6st + 4t²
= [(3s)/2]² + 2 × (3s/2) × (2t) + (2t)²
= [(3s)/2 + 2t]² [Using a² + 2ab + b² = (a + b)²]

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
Answer:
Grouping the terms as:
m²/9 + mk/3 + k²/4 + 2mn + 3nk + 9n²
We have:
m²/9 = (m/3)²
k²/4 = (k/2)²
9n² = (3n)²
Now checking the cross terms:
2 × (m/3) × (k/2) = mk/3
2 × (m/3) × (3n) = 2mn
2 × (k/2) × (3n) = 3nk
So, m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
= m²/9 + k²/4 + 9n² + + mk/3 + 3nk + 2mn [ Rearranging the terms]
= (m/3)² + (k/2)² + (3n)² + 2 × (m/3) × (k/2) + 2 × (m/3) × (3n) + 2 × (k/2) × (3n)
= (m/3 + k/2 + 3n)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

(iv) p²/16 – 2 + 16/p²
Answer:
Writing -2 as:
-2 = -2 × (p/4) × (4/p)
Now, p²/16 = (p/4)²
16/p² = (4/p)²
So, p²/16 – 2 + 16/p²
= (p/4)² – 2(p/4)(4/p) + (4/p)²
= (p/4 – 4/p)² [Using a² – 2ab + b² = (a – b)²]

(v) 9a² + 4b² + c² – 12ab + 6ac – 4bc
Answer:
We have:
9a² = (3a)²
4b² = (- 2b)² [Here 4b² = (- 2b)² as in two negative terms – 12ab and – 4bc, b is common]
c² = c²
Now, 9a² + 4b² + c² – 12ab + 6ac – 4bc
= (3a)² + (- 2b)² + c² + 2(3a)(- 2b) + 2(3a)(c) + 2(- 2b)(c)
= (3a – 2b + c)² [Using x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²]

(i) (p + 3q + 7r)²
Answer:
Here, a = p, b = 3q, c = 7r
Using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
So, (p + 3q + 7r)²
= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)²
Answer:
Write it as:
[3x + (−2y) + 4z]²
Here,
a = 3x, b = −2y, c = 4z
Using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
So, (3x − 2y + 4z)²
= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(3x)(4z)
= 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Answer:
To check whether this is an identity, expand the left-hand side.
First Part: (a + b − c)²
= a² + b² + c² + 2ab − 2ac − 2bc
Second Part: (a − b + c)²
= a² + b² + c² − 2ab + 2ac − 2bc
Third Part: (a − b − c)²
= a² + b² + c² − 2ab − 2ac + 2bc
Now adding all three:
LHS = (a² + b² + c² + 2ab − 2ac − 2bc)
+ (a² + b² + c² − 2ab + 2ac − 2bc)
+ (a² + b² + c² − 2ab − 2ac + 2bc)
= 3a² + 3b² + 3c² − 2ab − 2ac − 2bc
This is NOT equal to 2a² + 2b² + 2c²
Hence, the given statement is not true for all values of a, b and c.
So, it is not an identity.

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.4 Solutions

(i) s² − 11s + 24 = ( ____ )( ____ )
Answer:
We need two numbers whose:
Sum = 11 and Product = 24
These are 3 and 8.
So, s² − 11s + 24
= s² − 8s – 3s + 24
= s(s − 8) – 3(s − 8)
= (s − 3)(s − 8)
So, s² − 11s + 24 = = (s − 3)(s − 8)

(ii) ( ____ )(x + 1) = (3x² − 4x − 7)
Answer:
RHS: 3x² − 4x − 7
= 3x² − 7x + 3x − 7
= x(3x − 7) + 1(3x – 7)
= (3x − 7)(x + 1)
So, (3x – 7)(x + 1) = (3x² − 4x − 7)

(iii) 10x² − 11x − 6 = (2x − ____)( ____ + 2)
Answer:
LHS = 10x² − 11x − 6
= 10x² − 15x + 4x − 6
= 5x(2x − 3) + 2(2x − 3)
= (5x + 2)(2x − 3)
(2x − 3)(5x + 2)
So, 10x² − 11x − 6 = (2x − 3)(5x + 2)

(iv) 6x² + 7x + 2 = ( ____ )( ____ )
Answer:
= 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1)
So, 6x² + 7x + 2 = = (3x + 2)(2x + 1)

(i) (41)²
Answer:
41 = 40 + 1
= (40 + 1)²
= 40² + 2 × 40 × 1 + 1² [Using (a + b)² = a² + 2ab + b²]
= 1600 + 80 + 1
= 1681

(ii) (27)²
Answer:
27 = 30 − 3
So, (27)² = (30 − 3)²
= (30)² – 2 × 30 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]
= 900 − 180 + 9
= 729

(iii) (23 × 17)
Answer:
Use identity:
23 × 17 = (20 + 3)(20 − 3)
= 20² − 3² [Using (a + b)(a − b) = a² − b²]
= 400 − 9
= 391

(iv) (135)²
Answer:
135 = 100 + 35
So, (135)²
= (100 + 35)²
= (100)² + 2 × 100 × 35 + (35)² [Using (a + b)² = a² + 2ab + b²]
= 10000 + 7000 + 1225
= 18225

(v) (97)²
Answer:
97 = 100 − 3
So, (97)²
= (100 − 3)²
= (100)² – 2 × 100 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]
= 10000 − 600 + 9
= 9409

(vi) (18 × 29)
Answer:
18 × 29 =
= (20 − 2)(20 + 9)
= 20² + (− 2 + 9) × 20 − 18 [Using (x + a)(x + b) = x² + (a + b)x + a × b]
= 400 + 140 − 18
= 522

(vii) (34 × 43)
Answer:
= (38 − 4)(38 + 5)
= 38² + (− 4 + 5) × 38 − 20 [Using (x + a)(x + b) = x² + (a + b)x + a × b]
= 38² + 38 − 20
= 1444 + 38 − 20
= 1462

(viii) (205)²
Answer:
205 = 200 + 5
= (200 + 5)²
= (200)² + 2 × 200 × 5 + (5)² [Using (a + b)² = a² + 2ab + b²]
= 40000 + 2000 + 25
= 42025

(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc
Answer:
9a² + b² + 4c² − 6ab + 12ac − 4bc
= (3a)² + (b)² + (2c)² + 2(3a)(−b) + 2(−b)(2c) + 2(3a)(2c)
= (3a − b + 2c)² [Using x² + y² + z² + 2xy + 2xz + 2yz = (x + y + z)²]

(ii) 16s² + 25t² − 40st
Answer:
Now, 16s² + 25t² − 40st
= 16s² − 40st + 25t² [Rearranging the terms]
= (4s)² −2(4s)(5t) + (5t)²
= (4s − 5t)² [Using a² − 2ab + b² = (a − b)²]

(iii) r² − r − 42
Answer:
We need two numbers whose product = −42 and Sum = −1.
These numbers are −7 and 6.
So, r² − r − 42
= r² − 7r + 6r − 42
= r(r − 7) + 6(r − 7)
= (r − 7)(r + 6)

(iv) 49g² + 14gh + h²
Answer:
49g² + 14gh + h²
= (7g)² + 2(7g)(h) + (h)²
= (7g + h)² [Using a² + 2ab + b² = (a + b)²]

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw
Answer:
64u² + 121v² + 4w² − 176uv − 32uw + 44vw
= 64u² + 121v² + 4w² − 176uv + 44vw − 32uw [Rearranging the terms]
= (8u)² + (11v)² + (2w)² + 2(8u)(−11v) + 2(−11v)(−2w) + 2(8u)(−2w)
= (8u − 11v − 2w)² [Using x² + y² + z² + 2xy + 2xz + 2yz = = (x + y + z)²

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.5 Solutions

(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)
Answer:
First factorising the numerator:
3p² − 3pq − 18q²
= 3(p² − pq − 6q²)
= 3[p² − 3pq + 2pq − 6q²]
= 3[p(p − 3q) + 2q(p – 3q)
= 3(p − 3q)(p + 2q)

Now factorising the denominator:
p² + 3pq − 10q²
= p² + 5pq – 2pq − 10q²
= p(p + 5q) – 2q(p + 5q)
= (p + 5q)(p − 2q)

So, (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)
= 3(p − 3q)(p + 2q)/[(p + 5q)(p − 2q)]
No common factor cancels.

(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²)
Answer:
Factorising the numerator using identity:
n³ − 3n²m + 3nm² − m³
= (n − m)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]

Factorising the denominator:
5m² − 10mn + 5n²
= 5(m² − 2mn + n²)
= 5(m − n)² [Using a² – 2ab + b² = (a − b)²]

Now, (n − m)³/[5(m − n)²]
= −(m − n)³/[5(m − n)²] [Since, (n − m) = −(m − n), so, (n − m)³ = −(m − n)³]
= −(m − n)/5

(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)
Answer:
Factorising the numerator:
w³ − v³ + x³ + 3wvx
= w³ + (- v)³ + x³ – 3w(- v)x
= (w − v + x)(w² + v² + x² + wv + vx − wx)
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]

Now denominator:
w² + v² + x² − 2wv − 2vx + 2wx
= w² + (- v)² + x² + 2w(-v) + 2(−v)x + 2xw)
= (w − v + x)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

Therefore, (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)
= [(w − v + x)(w² + v² + x² + wv + vx − wx)]/(w − v + x)²
= (w² + x² + v² − wx + vx − wv) / (w + x − v) [Canceling the common factor]

(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)
Answer:
Factor numerator:
4y² − 20yz + 25z²
= (2y)² − 2(2y)(5z) + (5z)²
= (2y − 5z)² [Using a² – 2ab + b² = (a − b)²]

Factorising the denominator:
25z² − 4y²
= (5z − 2y)(5z + 2y) [Using a² – b² = (a − b)(a + b)]

So, (4y² − 20yz + 25z²)/(25z² − 4y²)
= (5z − 2y)²/[(5z − 2y)(5z + 2y)] [Since (2y − 5z)² = (−1)²(5z − 2y)² = (5z − 2y)²]
= (5z − 2y)/(5z + 2y)

(v) [(x² + x − 6)(x² − 7x + 12)]/[(x² − 6x + 8)(x² − 9)]
Answer:
Factorising each polynomial:
x² + x − 6 = x² + 3x – 2x − 6 = x(x + 3) – 2(x + 3) = (x + 3)(x − 2)
x² − 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x − 3)(x − 4)
x² − 6x + 8 = x² – 4x – 2x + 8 = x(x – 4) – 2(x – 4) = (x − 2)(x − 4)
x² − 9 = x² − 3 = (x − 3)(x + 3)
Substituting each polynomial, we get:
[(x + 3)(x − 2)][(x − 3)(x − 4)]/[(x − 2)(x − 4)][(x − 3)(x + 3)]
= 1 [Since all factors cancel]

(vi) (p⁴ − 16) / (p² − 4p + 4)
Answer:
Factorising numerator:
p⁴ − 16
= (p² − 4)(p² + 4) [Using a² − b² = (a + b)(a − b)]
= (p − 2)(p + 2)(p² + 4) [Again using a² − b² = (a + b)(a − b)]
Factorising denominator:
p² − 4p + 4
= p² − 2(p)(2) + 2²
= (p − 2)² [Using a² – 2ab + b² = (a – b)²]
So, [(p − 2)(p + 2)(p² + 4)]/(p − 2)²
= (p + 2)(p² + 4) / (p − 2) [Cancelling common factor (p − 2)].

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

(i) (−3x + 4)²
Answer:
(−3x + 4)²
= (−3x)² + 2(−3x)(4) + 4² [Using the identity (a + b)² = a² + 2ab + b²]
= 9x² − 24x + 16

(ii) (2s + 7)(2s − 7)
Answer:
(2s + 7)(2s − 7)
= (2s)² − 7² [Using the identity (a + b)(a − b) = a² − b²]
= 4s² − 49

(iii) (p² + 1/2)(p² − 1/2)
Answer:
(p² + 1/2)(p² − 1/2)
= (p²)² − (1/2)² [Using the identity (a + b)(a − b) = a² − b²]
= p⁴ − 1/4

(iv) (2n + 7)(2n − 7)
Answer:
(2n + 7)(2n − 7)
= (2n)² − 7² [Using the identity (a + b)(a − b) = a² − b²]
= 4n² − 49

(v) (s − 2t)(s² + 2st + 4t²)
Answer:
(s − 2t)(s² + 2st + 4t²)
= s³ − (2t)³ [Using the identity (a − b)(a² + ab + b²) = a³ − b³]
= s³ − 8t³

(vi) [1/(2r) − 4r]²
Answer:
[1/(2r) − 4r]²
= (1/(2r))² − 2(1/(2r))(4r) + (4r)² [Using the identity (a − b)² = a² − 2ab + b²]
= 1/(4r²) − 4 + 16r²

(vii) (−3m + 4k − l)²
Answer:
So, (−3m + 4k − l)²
= (−3m)² + (4k)² + (−l)² + 2(−3m)(4k) + 2(4k)(−l) + 2(−3m)(−l)
[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= 9m² + 16k² + l² − 24mk − 8kl + 6ml

(viii) (x − 1/3 y)³
Answer:
(x − 1/3 y)³
= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³
[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]
= x³ − x²y + 3x(y²/9) − y³/27
= x³ − x²y + (1/3)xy² − y³/27

(ix) (7/2 k − 2/3 m)³
Answer:
[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]
Here, a = 7k/2, b = 2m/3
So, a³ = (7k/2)³ = 343k³/8
3a²b = 3 × (49k²/4) × (2m/3) = 49k²m/2
3ab² = 3 × (7k/2) × (4m²/9) = 14km²/3
b³ = (2m/3)³ = 8m³/27
Therefore, (7/2 k − 2/3 m)³
= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27.

(i) 17 × 21
Answer:
Using identity: (a − b)(a + b) = a² − b²
We have 17 × 21 = (19 − 2)(19 + 2)
= 19² − 2²
= 361 − 4
= 357

(ii) 104 × 96
Answer:
Using identity: (a + b)(a − b) = a² − b²
We have 104 × 96 = (100 + 4)(100 − 4)
= 100² − 4²
= 10000 − 16
= 9984

(iii) 24 × 16
Answer:
Using identity: (a + b)(a − b) = a² − b²
We have 24 × 16 = (20 + 4)(20 − 4)
= 20² − 4²
= 400 − 16
= 384

(iv) 147³
Answer:
We know that 147 = 150 − 3
So, 147³ = (150 − 3)³
= 150³ − 3 × 150² × 3 + 3 × 150 × 3² − 3³ [Using identity (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 3375000 − 202500 + 4050 − 27
= 3176523

(v) 199³
Answer:
199 = 200 − 1
So, 199³ = (200 − 1)³
= 200³ − 3 × 200² × 1 + 3 × 200 × 1² − 1 [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 8000000 − 120000 + 600 − 1
= 7880599

(vi) 127³
Answer:
Here, 127 = 130 − 3
So, 127³ = (130 − 3)³
= 130³ − 3 × 130² × 3 + 3 × 130 × 3² − 3³ [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 2197000 − 152100 + 3510 − 27
= 2048383

(vii) (−107)³
Answer:
(−107)³ = −(107³)
Now, 107 = 100 + 7
107³ = (100 + 7)³
= 100³ + 3 × 100² × 7 + 3 × 100 × 7² + 7³ [Using identity: (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 210000 + 14700 + 343
= 1225043
So, (−107)³ = −1225043

(viii) (−299)³
Answer:
(−299)³ = −(299³)
Here, 299 = 300 − 1
So, 299³ = (300 − 1)³
= 300³ − 3 × 300² × 1 + 3 × 300 × 1² − 1 [Using identity: (a − b)³ = a³ − 3a²b + 3ab² − b³]
= 27000000 − 270000 + 900 − 1
= 26730900 − 1
= 26730899
So, (−299)³ = −26730899

(i) 4y² + 1 + 1/(16y²)
Answer:
Here, we have 4y² = (2y)²
1/(16y²) = (1/4y)²
and 2 × 2y × 1/(4y) = 1
So, this is of the form: a² + 2ab + b² = (a + b)²
Therefore, 4y² + 1 + 1/(16y²)
= (2y + 1/(4y))²

(ii) 9m² − 1/(25n²)
Answer:
9m² − 1/(25n²)
= (3m)² − (1/5n)²
= (3m + 1/5n)(3m − 1/5n) [Using a² − b² = (a + b)(a − b)]

(iii) 27b³ − 1/(64b³)
Answer:
27b³ − 1/(64b³)
= [3b − 1/(4b)][(3b)² + (3b)(1/4b) + (1/4b)²] [Using a³ − b³ = (a − b)(a² + ab + b²)]
= (3b − 1/4b)[9b² + 3/4 + 1/16b²]

(iv) x² + 5x/6 + 1/6
Answer:
We need two numbers whose sum is 5/6 and product is 1/6.
These numbers are 1/2 and 1/3.
So, x² + 5x/6 + 1/6
= x² + x/2 + x/3 + 1/6 [Splitting the middle term]
= x(x + 1/2) + 1/3(x + 1/2)
= (x + 1/3)(x + 1/2)

(v) 27u³ − 1/125 − 27u²/5 + 9u/25
Answer:
Given expression:
27u³ − 1/125 − 27u²/5 + 9u/25
= 27u³ − 27u²/5 + 9u/25 − 1/125 [Rearranging the terms]
= (3u)³ − 3(3u)²(1/5) + 3(3u)(1/5)² − (1/5)³
= (3u − 1/5)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]

(vi) 64y³ + 1/125 z³
Answer:
64y³ + 1/125 z³
= (4y)³ + (z/5)³
= (4y + z/5)[(4y)² − (4y)(z/5) + (z/5)²] [Using a³ + b³ = (a + b)(a² − ab + b²)]
= (4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr
Answer:
p³ + 27q³ + r³ − 9pqr
= p³ + (3q)³ + r³ − 3(p)(3q)(r)
= (p + 3q + r)[p² + (3q)² + r² − p(3q) − (3q)r − pr)]
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]
= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)

(viii) 9m² − 12m + 4
Answer:
9m² − 12m + 4
= (3m)² − 2(3m)(2) + (2)²
= (3m – 2)² [Using a² – 2ab + b² = (a – b)²]

(ix) 9x³ − 8/3 y³ + z³/3 + 6xyz
Answer:
9x³ − 8/3 y³ + z³/3 + 6xyz
= (1/3)[9x³ − 8y³ + z³ + 18xyz]
= (1/3)[(3x)³ + (- 2y)³ + z³ – 3(3x)(-2y)z]
= (1/3)(3x – 2y + z)[(3x)² + (-2y)² + z² − (3x)(-2y) − (-2y)(z) − (z)(3x)]
[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]
= (1/3)(3x – 2y + z)[9x² + 4y² + z² + 6xy + 2yz − 3zx]

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
Answer:
4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= 4x² + 9y² + 36z² + 24xy + 36yz + 12xz
= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

(xi) 27u³ − 1/216 − 9u²/2 + u/4
Answer:
27u³ − 1/216 − 9u²/2 + u/4
= 27u³ − 9u²/2 + u/4 − 1/216 [Rearranging the terms]
= (3u)³ − 3(3u)²(1/6) + 3(3u)(1/6)² − (1/6)³
= (3u − 1/6)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³].

Note: Assume that the denominators are not equal to 0.
(i) (4x² + 4x + 1)/(4x² − 1)
Answer:
Factorising numerator:
4x² + 4x + 1
= (2x)² + 2(2x)(1) + 1²
= (2x + 1)² [Using a² + 2ab + b² = (a + b)²]
Factorising denominator:
4x² − 1 = (2x + 1)(2x − 1)
Now the expression: (4x² + 4x + 1)/(4x² − 1)
= (2x + 1)²/[(2x + 1)(2x − 1)]
= (2x + 1)/(2x − 1)

(ii) 9(3a³ − 24b³)/(9a² − 36b²)
Answer:
First simplify:
9(3a³ − 24b³) = 27(a³ − 8b³)
and 9a² − 36b² = 9(a² − 4b²)
So, 9(3a³ − 24b³)/(9a² − 36b²)
= 3(a³ − 8b³)/(a² − 4b²)
= 3(a − 2b)(a² + 2ab + 4b²)/(a² − 4b²) [Since a³ − 8b³ = a³ − (2b)³ = (a − 2b)(a² + 2ab + 4b²)]
= 3(a − 2b)(a² + 2ab + 4b²)/(a − 2b)(a + 2b) [Since a² − 4b² = (a − 2b)(a + 2b)]
= 3(a² + 2ab + 4b²)/(a + 2b) [Cancelling the common factor (a − 2b)]

(iii) (s³ + 125t³)/(s² − 2st − 35t²)
Answer:
Factorising numerator:
s³ + 125t³ = s³ + (5t)³
= (s + 5t)(s² − 5st + 25t²) [Using a³ + b³ = (a + b)(a² – ab + b²)]
Factorising denominator:
s² − 2st − 35t²
s² − 7st + 5st − 35t²
= s(s – 7t) + 5t(s − 7t)
= (s + 5t)(s − 7t)

Now the given expression:
(s³ + 125t³)/(s² − 2st − 35t²)
= (s + 5t)(s² − 5st + 25t²)/(s + 5t)(s − 7t)
= (s² − 5st + 25t²)/(s − 7t) [Cancelling common factor (s + 5t)].

(i) 25a² − 30ab + 9b²
Answer:
This is a perfect square:
25a² − 30ab + 9b²
= (5a)² − 2(5a)(3b) + (3b)²
= (5a − 3b)² [Using a² – 2ab + b² = (a – b)²]
So possible length and breadth are (5a − 3b) and (5a − 3b).

(ii) 36s² − 49t²
Answer:
36s² − 49t²
= (6s)² − (7t)²
= (6s + 7t)(6s − 7t) [Using a² – b² = (a + b)(a – b)]
So possible length and breadth are (6s + 7t) and (6s − 7t).

(i) 6a² − 24b²
Answer:
Take common factor:
6a² − 24b²
= 6(a² − 4b²)
= 6[a² − (2b)²]
= 6(a + 2b)(a − 2b) [Using a² – b² = (a + b)(a – b)]
So possible dimensions are 6, (a + 2b) and (a − 2b).

(ii) 3ps² − 15ps + 12p
Answer:
3ps² − 15ps + 12p
= 3p(s² − 5s + 4) [Taking common factor]
= 3p(s² − 4s – 1s + 4)
= 3p[s(s − 4) – 1(s − 4)]
= 3p(s − 1)(s − 4)
So, the possible dimensions are 3p, (s − 1) and (s − 4).

Answer:
Side of playground = 40 m
Since a path of width s metres is made all around the outside, the side of the outer square becomes:
40 + 2s
Area of outer square = (40 + 2s)²
Area of playground = 40² = 1600
Therefore, area of the path
= (40 + 2s)² − 1600
= (1600 + 160s + 4s²) − 1600
= 4s² + 160s
Hence, the required expression: Area of path = 4s² + 160s square metres.

Answer:
Let the number be x.
Then, according to question: x + 1/x = 10/3
Multiply both sides by 3x:
3x² + 3 = 10x
⇒ 3x² − 10x + 3 = 0
⇒ 3x² − 9x − x + 3 = 0
⇒ 3x(x − 3) − 1(x − 3) = 0
⇒ (3x − 1)(x − 3) = 0
So, 3x − 1 = 0 or x − 3 = 0
Hence, x = 1/3 or x = 3
Therefore, the number is 3 or 1/3.

Answer:
Area of pool = 2x² + 7x + 3
Width = 2x + 1
Length = Area/Width
= (2x² + 7x + 3)/(2x + 1)
= (2x² + 6x + x + 3)/(2x + 1)
= [2x(x + 3) + 1(x + 3)]/(2x + 1)
= (2x + 1)(x + 3)/(2x + 1)
= x + 3
Therefore, the length of the pool is x + 3 hastas.

Answer:
Since x − 2 and x − 1/2 are factors, the quadratic polynomial can be written as:
px² + 5x + r = k(x − 2)(x − 1/2)
⇒ (x − 2)(x − 1/2) = k[x² − (5/2)x + 1]
⇒ x² − (1/2)x − 2x + 1 = k[x² − (5/2)x + 1]
⇒ x² − (5/2)x + 1 = k[x² − (5/2)x + 1]
Comparing the coefficients on both sides, we get:
Coefficient of x² gives: k = p
Constant term gives: k = r
Therefore, p = r
Hence proved.

Answer:
Given:
a + b + c = 5
ab + bc + ca = 10

First finding a² + b² + c² using:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
So, 5² = a² + b² + c² + 2(10)
⇒ 25 = a² + b² + c² + 20
⇒ a² + b² + c² = 5
Now, a² + b² + c² − ab − bc − ca = 5 − 10 = −5

Using the identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
a³ + b³ + c³ − 3abc
= (5)(−5)
= −25
Hence proved a³ + b³ + c³ − 3abc = −25.

Answer:
We have: n³ − n
= n(n² − 1)
= n(n − 1)(n + 1)
So, n³ − n = n(n − 1)(n + 1)
Thus, n³ − n is the product of three consecutive natural numbers: (n − 1), n, (n + 1)
Among any three consecutive natural numbers:

• one is always divisible by 3
• at least one is always even, so divisible by 2

Therefore, their product is always divisible by: 2 × 3 = 6
Hence, n³ − n is always divisible by 6 for all natural numbers n.

(i) x³ + y³ − 12xy + 64, when x + y = −4
Answer:
Since x + y = −4, we have:
(x + y)³ = (−4)³ = −64
Using identity: (x + y)³ = x³ + y³ + 3xy(x + y)
So,
−64 = x³ + y³ + 3xy(−4)
⇒ −64 = x³ + y³ − 12xy

Therefore, x³ + y³ − 12xy = −64
Now, x³ + y³ − 12xy + 64 = −64 + 64 = 0
Hence, the value is 0.

(ii) x³ − 8y³ − 36xy − 216, when x − 2y + 6 = 0
Answer:
Given: x − 2y + 6 = 0
So, x − 2y = −6
Now use the identity: (a − b)³ = a³ − 3a²b + 3ab² − b³
We have: x³ − 8y³ − 36xy − 216
= x³ − 8y³ + 3(x)(−2y)(−6) + (−6)³
So it matches: a³ + b³ + c³ − 3abc
with a = x, b = −2y, c = −6
Using identity: a³ + b³ + c³ − 3abc
= (a + b + c)(a² + b² + c² − ab − bc − ca)
Thus, x³ − 8y³ − 36xy − 216
= x³ − 8y³ − 216 − 36xy
= x³ + (-2y)³ + (-6) − 3(x)(-2y)(-6)
= (x − 2y − 6)[x² + 4y² + 36² + 2xy − 12y + 18x]
But from the question:
x − 2y + 6 = 0, which gives x − 2y = −6, so x − 2y − 6 = −12, not zero.
So this expression does not become 0 under the given condition as written.
Using x = 2y − 6 from the condition, Substitute into the expression:
x³ − 8y³ − 36xy − 216
= (2y − 6)³ − 8y³ − 36(2y − 6)y − 216
= (8y³ − 72y² + 216y − 216) − 8y³ − 72y² + 216y − 216
= −144y² + 432y − 432
= −144(y² − 3y + 3)
Therefore, under the given condition, the value is −144(y² − 3y + 3).
So, the expression does not reduce to a constant unless there is a typo in the question.

Class 9 Maths Ganita Manjari Chapter 4 List of Formulae Used

The following is the complete list of identities studied in this chapter:

What is an Algebraic Identity?
An algebraic identity is an equation that is true for all values of the variables in it. For example, (x + y)² = x² + 2xy + y² is true whether x and y are positive, negative, rational, or any real number. This is the fundamental difference between an identity and an ordinary equation – an equation like x² − 1 = 24 is satisfied only for specific values (x = 5 or x = −5), making it an equation but not an identity.
Difference Between Equation and Identity:

• An equation is true only for specific values of the variable.
• An identity is true for all values of the variable.

Geometric Visualisation:
The chapter uses a particularly beautiful approach – geometric squares and rectangles – to justify why identities like (a + b)² = a² + 2ab + b² work. A square of side (a + b) units is divided into one a² square, one b² square, and two ab rectangles, confirming the identity visually.
Historical Note – Śhrīdharāchārya’s Method (750 CE):
The textbook highlights that the identity a² = (a + b)(a − b) + b² was used by the Indian mathematician Śhrīdharāchārya in 750 CE as a method to quickly compute squares of numbers — a proud reminder of India’s mathematical heritage.

• Exercise Set 4.1: Expansion of expressions using (a + b)² = a² + 2ab + b²; finding squares of numbers like (64)², (105)², (205)².
• Exercise Set 4.2: Factorisation using perfect square identities; finding values using (a − b)² = a² − 2ab + b².
• Exercise Set 4.3: Squaring three-digit numbers using (a + b)² and (a + b + c)²; factorisation using multiple identities.
• Exercise Set 4.4: Completing factorisation blanks; products using identities; factorisation of expressions involving three-variable identities.
• Exercise Set 4.5: Simplification of rational algebraic expressions using factorisation.
• End-of-Chapter Exercises: A comprehensive set of problems covering all identities, word problems (rectangular pool, playground path, algebra tile arrangements), advanced starred questions on divisibility and proofs.

The chapter includes several starred problems meant for advanced students:

• Q10: If both (x − 2) and (x − ½) are factors of px² + 5x + r, prove that p = r.
• Q11: If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.
• Q12: Prove by factorisation that n³ − n is always divisible by 6 for all natural numbers n.
• Q13(i): Find x³ + y³ − 12xy + 64 when x + y = −4.
• Q13(ii): Find x³ − 8y³ − 36xy − 216 when x = 2y + 6.

These questions require combining multiple identities and are excellent practice for competitive examinations and Olympiad preparation.