Question

Factorise:

729a⁶ – b⁶ 

Answer 1

Given, 729a⁶ – b⁶ 

Now, 729a⁶ – b⁶ can be written as (27a³)² – (b³)²

Using the identity, 

a² – b² = (a + b) (a – b), we get,

(27a³)² – (b³)² = (27a³ + b³) (27a³ – b³)

Now, [(3a)³ + (b)³] [(3a)³ – (b)³]

⇒ (3a + b) (9a² – 3ab + b²) (3a – b) (9a² + 3ab + b²)

Hence, the required is (3a + b) (3a – b) (9a² – 3ab + b²) (9a² + 3ab + b²)