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).

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