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# Factorise: 27x³ + y³ + z³ - 9xyz

**Solution:**

We will use the algebraic identity: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

The given expression 27x³ + y³ + z³ - 9xyz can be written as (3x)³ +(y)³ +(z)³ - 3(3x)(y)(z)

By using the identity x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

We can write: (3x)³ + (y)³ + (z)² - 3(3x)(y)(z) = (3x + y + z)[(3x)² + ( y)² + (z)² - (3x)( y) - yz - (z)(3x)]

Hence, 27x³ + y³ + z³ - 9xyz = (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)

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

**Video Solution:**

## Factorise: 27x³ + y³ + z³ - 9xyz

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 11:

**Summary:**

The factorized form of the expression 27x³ + y³ + z³ - 9xyz is (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx).

**☛ Related Questions:**

- 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)
- Factorise each of the following:i) 27y³ + 125z³ii) 64m³ - 343n³ = (4m)³ - (7n)³[Hint: See Question 9.]

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