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# Factorise each of the following:

i) 27y³ + 125z³ ii) 64m³ - 343n³

[Hint: See Question 9]

**Solution:**

i) 27y³ + 125z³ = (3y)³ + (5z)³

Using factorization: (x³ + y³) = (x + y)(x² - xy + y²)

Here, x = 3y and y = 5z

(3y)³ + (5z)³ = (3y + 5z)[(3y)² - (3y)(5z) + (5z)²]

Thus, 27y³ + 125z³ = (3y + 5z)(9y² - 15yz + 25z²)

ii) 64m³ - 343n³ = (4m)³ - (7n)³

Using factorization: (x³ - y³) = (x - y)(x² + xy + y²)

Here, x = 4m and y = 7n

(4m)³ - (7n)³ = (4m - 7n)[(4m)² + (4m)(7n) + (7n)²]

Thus, 64m³ - 343n³ = (4m - 7n)(16m² + 28mn + 49n²)

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Factorise each of the following: i) 27y³ + 125z³ ii) 64m³ - 343n³ [Hint: See Question 9]

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 10

**Summary:**

The factorized form of each of the following 27y³ + 125z³ and 64m³ − 343n³ are (3y + 5z)(9y² − 15yz + 25z²) and (4m −7n)(16m² + 28mn + 49n²) respectively.

**☛ Related Questions:**

- Write the following cubes in expanded form:i) (2x +1)³ii) (2a - 3b)³iii) (3/2x + 1)³iv) (x - 2/3y)³
- Evaluate the following using suitable identities:(i) (99)3(ii) (102)3(iii) (998)3
- Factorise each of the following:i) 8a³ + b³ + 12a²b + 6ab²ii) 8a³ - b³ - 12a²b + 6ab²iii) 27 - 125a³ - 135a + 225a²iv) 64a³ - 27b³ - 144a²b + 108ab²v) 27p³ - 1/216 - 9/2p² + 1/4p
- Verify:i) (x³ + y³) = (x + y)(x² - xy + y²)ii) (x³ - y³) = (x - y)(x² + xy + y²)

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