OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Ex 4(H)

Students often turn to OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(H) to clarify doubts and improve problem-solving skills.

S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(H)

Factorise the following :

Question 1.
a³ + 1
Solution:
a³ + 1 = (x)³ + (1)³
= (x + 1) (x² – x + 1)

Question 2.
x³ + 8
Solution:
x³ + 8 = (x)³ + (2)³
= (x + 2) [(x)² – x × 2 + (2)²]
= (x + 2) (x² – 2x + 4)

Question 3.
8x³ + 1
Solution:
8x³ + 1 = (2x)³ + (1)³
= (2x + 1) [(2x)² – 2x × 1 + (1)²]
= (2x + 1) (4x² – 2x + 1)

Question 4.
x³ – 27
Solution:
x³ – 27 = (x)³ – (3)³
= (x – 3) [(x)² + x × 3 + (3)²]
= (x – 3) (x² + 3x + 9)

Question 5.
a³ – 8
Solution:
a³ – 8 = (a)³ – (2)³
= (a – 2) [(a)² + a x 2 + (2)²]
= (a – 2) (a² + 2a + 4)

Question 6.
27m³ – 8
Solution:
27m³ – 8 = (3m)³ – (2)³
= (3m – 2) [(3m)² + 3m x 2 + (2)²]
= (3m – 2) (9m)² + 6m + 4)

Question 7.
x³+ 64
Solution:
x³ + 64 = (x)³ + (4)³
= (x + 4) [(x)² – x × 4 + (4)²]
= (x + 4) (x² – 4x + 16)

Question 8.
8a³ – b⁶
Solution:
8a³ – b⁶ = (2a)³ – (b²)²
= (2a – b²) [(2a)² + 2a x b² + (b²)²]
= (2a – b²) (4a² + 2ab² + b⁴)

Question 9.
x⁶ + 8b³
Solution:
x⁶ + 8b³ = (x²)³ + (2b)³
= (x² + 2b) [(x²)² – x² x 2b + (2b)²]
= (x² + 2b) (x⁴ – 2x²b + 4b²)

Question 10.
8a³ + 21b³
Solution:
8a³ + 21b³ = (2a)³ + (3b)³
= (2a + 3b) [(2a)² – 2a x 3b + (3b)²]
= (2a + 3b) (4a² – 6ab + 9b²)

Question 11.
27x³ – 8y³
Solution:
27x³ – 8y³ = (3x)³ – (2y)³
= (3x – 2y) [(3x)² + 3x × 2y + (2y)²]
= (3x – 2y) (9x² + 6xy + 4y²)

Question 12.
128x³ + 2
Solution:
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)

Question 13.
Factorise : 8x³ – \(\frac { 1 }{ 27 }\)y³
Solution:

Question 14.
343x³y + 512y⁴
Solution:
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)

Question 15.
(2a + b)³ + (a + 2b)³
Solution:
(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)

Question 16.
27 (m + 2n)³ + (m – 6n)³
Solution:
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]

Question 17.
8 (a + b)³ – 21c³
Solution:
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)

Question 18.
x⁶ – 1
Solution:
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)

Question 19.
64a⁶ – b⁶
Solution:
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)

Question 20.
a³ – b³ + 4 (a – b)
Solution:
a³ – b³ + 4 (a – b)
= (a – b) (a² + ab + b²) + 4 (a – b)
= (a – b) (a² + ab + b³ + 4)

Question 21.
\(x^3-\frac{1}{x^3}-6 x+\frac{6}{x}\)
Solution:

Question 22.
64a³ + 125b³ + 12a²b + 15ab²
Solution:
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)

Question 23.
375 (a – b)³+ 3
Solution:
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)

Question 24.
If a⁴ + b⁴ = a²b², Show that a⁶ + b⁶ = 0
Solution:
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.