Factorisation Class 9 OP Malhotra Exe-4H ICSE Maths Solutions Ch-4. We Provide Step by Step Solutions / Answer of Questions. How to factorize (a)³ + (b)³ expression. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Factorisation Class 9 OP Malhotra Exe-4H ICSE Maths Solutions Ch-4
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | OP Malhotra |
| Exe-4H | How to factorize (a)³ + (b)³ |
| Edition | 2025-2026 |
Exercise- 4H
Factorisation Class 9 OP Malhotra Exe-4H ICSE Maths Solutions Ch-4.
Que-1: a³ + 1
Sol: a³ + 1
= (x)³ + (1)³
= (x + 1) (x² – x + 1)
Que-2: x³ + 8
Sol: x³ + 8
= (x)³ + (2)³
= (x + 2) [(x)² – x × 2 + (2)²]
= (x + 2) (x² – 2x + 4)
Que-3: 8x³ + 1
Sol: 8x³ + 1
= (2x)³ + (1)³
= (2x + 1) [(2x)² – 2x × 1 + (1)²]
= (2x + 1) (4x² – 2x + 1)
Que-4: x³ – 27
Sol: x³ – 27
= (x)³ – (3)³
= (x – 3) [(x)² + x × 3 + (3)²]
= (x – 3) (x² + 3x + 9)
Que-5: a³ – 8
Sol: a³ – 8
= (a)³ – (2)³
= (a – 2) [(a)² + a x 2 + (2)²]
= (a – 2) (a² + 2a + 4)
Que-6: 27m³ – 8
Sol: 27m³ – 8
= (3m)³ – (2)³
= (3m – 2) [(3m)² + 3m x 2 + (2)²]
= (3m – 2) (9m)² + 6m + 4)
Que-7: x³+ 64
Sol: x³ + 64
= (x)³ + (4)³
= (x + 4) [(x)² – x × 4 + (4)²]
= (x + 4) (x² – 4x + 16)
Que-8: 8a³ – b6
Sol: 8a³ – b6
= (2a)³ – (b²)²
= (2a – b²) [(2a)² + 2a x b² + (b²)²]
= (2a – b²) (4a² + 2ab² + b4)
Que-9: x6 + 8b³
Sol: x6 + 8b³
= (x²)³ + (2b)³
= (x² + 2b) [(x²)² – x² x 2b + (2b)²]
= (x² + 2b) (x4 – 2x²b + 4b²)
Que-10: 8a³ + 21b³
Sol: 8a³ + 21b³
= (2a)³ + (3b)³
= (2a + 3b) [(2a)² – 2a x 3b + (3b)²]
= (2a + 3b) (4a² – 6ab + 9b²)
Que-11: 27x³ – 8y³
Sol: 27x³ – 8y³
= (3x)³ – (2y)³
= (3x – 2y) [(3x)² + 3x × 2y + (2y)²]
= (3x – 2y) (9x² + 6xy + 4y²)
Que-12: 128x³ + 2
Sol: 128x³ + 2
= 2 (64x³ + 1)
= 2 [(4x)³ + (1)³]
= 2 (4x + 1) [(4x)² – 4x × 1 + (1)²]
= 2 (4x + 1) (16x² – 4x + 1)
Que-13: Factorise : 8x³ – (1/27) y³
Sol: 8x³ – (1/27) y³
= (2x)³ – (1/3y)³
= (2x – (1/3y)) [(2x)²+(1/3y)²+{(2x) (1/3y)}]
= (2x – (1/3y)) (4x²+(1/9y²)+(2/3xy))
Que-14: 343x³y + 512y4
Sol: 343x³y + 512y4 = y (343x³ + 512y³)
= y[(7x)³ + (8y)³]
= y [(7x + 8y) [(7x)² – 7x × 8y + (8y)²]
= y (7x + 8y) (49x² – 56xy + 64y²)
Que-15: (2a + b)³ + (a + 2b)³
Sol: (2a + b)³ + (a + 2b)³
Let 2a + b = x and a + 2b = y,
then x³ + y³ = (x + y) [x² – xy + y²]
Substituting the value of x and y
(2 a + b)³ + (a + 2b)³ = (2 a + b + a + 2b)
[(2a + b)² – (2a + b) (a + 2b) + (a + 2b)²]
= (3a + 3b) [4a² + 4ab + b²- 2a² – 4ab – ab – 2b² + a² + 4b² + 4ab]
= 3 (a + b) [3a² + 3ab + 3b²]
= 3 (a + b) 3 (a² + ab + b²)
= 9 (a + b) (a² + ab + b²)
Que-16: 27 (m + 2n)³ + (m – 6n)³
Sol: 27 (m + 2n)³ + (m – 6n)³
Let m + 2n = a and m – 6n = b, then
27a³ + b³ = (3a)³ + (b)³ = (3a + b) [(3a)² – 3a x + (b)²]
= (3a + b) [9a² – 3ab + b²]
Substituting the value of a and b, we get
27 (m + 2M)³ + (m – 6n)³
= [3 (m + 2m) + m – 6m] [9 (m + 2m)² – 3 (m + 2n) (m – 6n) + (m – 6n)²]
= [3m² + 6n + m – 6n] [9 (m² + 4n² + 4mm) – 3 (m² – 6mn + 2nm – 12n²) + m² + 36n² – 12mn]
= 4m [9m² + 36m² + 36mn – 3m² + 18mn – 6mn + 36n² + m² + 36n² – 12mn]
= 4m [7m² + 36mn + 108n²]
Que-17: 8 (a + b)³ – 21c³
Sol: 8 (a + b)³ – 21c³
[2 (a + b)]³ – (3c)³
= [2 (a + b) – 3c] [{2 (a + b)}² + 2(a + b) x 3c + (3c)²]
= (2a + 2b – 3c) [4 (a² + b² + 2ab) + 6ac + 6bc + 9c²]
= (2a + 2b – 3c) [4a² + 4b² + Sab + 6ac + 6bc + 9c²]
= (2a + 2b – 3c) (4a² + 4b² + 9c² + 8ab + 6ac + 6bc)
Que-18: x6 – 1
Sol: x6 – 1
= (x³)² – (1)² {a² – b² = (a + b) (a – b)}
= (x³ + 1) (x³ – 1)
= [(x)³ + (1)³] [(x)³ – (1)³]
= (x + 1) (x² – x + 1) (x – 1) (x² + x + 1)
= (x + 1) (x – 1) (x² – x + 1) (x² + x + 1)
Que-19: 64a6 – b6
Sol: 64a6 – b6
= (8a³)² – (b³)²
= (8a³ + b³) (8a³ – b³) {a² -b² = (a + b)(a- b)}
= [(2a)³ + (b)³] [(2a)³ – (b)³]
= (2a + b) [(2a)² – 2a x b + (b)²] (2a – b) [(2a)² + 2a x b + b²]
= (2a + b) (4a² – 2ab + b²) (2a – b) (4a² + 2ab + b²)
= (2a + b) (2a – b) (4a² + 2ab + b²) (4a² – 2ab + b²)
Que-20: a³ – b³ + 4 (a – b)
Sol: a³ – b³ + 4 (a – b)
= (a – b) (a² + ab + b²) + 4 (a – b)
= (a – b) (a² + ab + b³ + 4)
Que-21: x³ − (1/x³) − 6x + (6/x)
Sol: x³ − (1/x³) − 6x + (6/x)
{x-(1/x)} {x²+1+(1/x²)} – 6 {x-(1/x}
{x-(1/x)} {x²+1+(1/x²)-6}
{x-(1/x)} {x²-5+(1/x²)}
Que-22: 64a³ + 125b³ + 12a²b + 15ab²
Sol: 64a³ + 125b³ + 12a²b² + 15ab²
= [(4a)³ + (5b)³] + 3ab (4a + 5b)
= (4a + 5b) [(4a)² – 4a x 5b + (5b)²] + 3ab (4a + 5b)
= (4a + 5b) (16a² – 20ab + 25b²) + 3ab (4a + 5 b)
= (4a + 5b) (16a² – 20ab + 25b² + 3ab)
= (4a + 5b) (16a² – 17ab + 25b²)
Que-23: 375 (a – b)³+ 3
Sol: 375 (a – b)³ + 3 = 3 [125 (a – b)³ + 1]
= 3 [{5 (a – b)}³ + (1)³]
= 3 [{5(a – b)+ 1} {(5 (a – b)}² – 5 (a – b) x 1 + (1)²]
= 3 [(5a – 5b + 1) (25 (a² + b² – 2ab) – 5a + 5b + 1)]
= 3 (5a – 5b + 1) (25a² + 25b² – 50ab – 5a + 5b + 1)
Que-24: If a4 + b4 = a²b², Show that a6 + b6 = 0
Sol: a4 + b4 = a²b²
L.H.S. = a6 + b6 = (a²)³ + (b²)³
= (a² + b²) [a4 – a²b² + b4]
= (a² + b²) (a4 + b4 – a²b²)
= (a² + b²) (a²b² – a²b²) (∵ a4 + b4 = a²b² given)
= (a² + b²) x 0 = 0 = R.H.S.
— : End of Factorisation Class 9 OP Malhotra Exe-4H ICSE Maths Solutions Ch-4 :–
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