Mark the tick against the correct answer in the following:

To find: Value of

We have,

Applying

⇒

Applying

⇒

We know, x³ – y³ = ( x – y) (x²+xy+y²)

⇒

Taking (b-a) common from C₂

⇒

Taking (c-a) common from C₂

⇒

Expanding along C₁

⇒ (b - a) (c – a)[1{(1)(c² + ca + a²) – (b² + ab + a²)(1)}]

⇒ (b - a) (c – a)[c² + ca + a² – b² - ab - a²]

⇒ (b - a) (c – a)[c² – b² + ca - ab]

⇒ (b - a) (c – a)[(c – b) (c + b) + a(c – b)]

⇒ (b - a) (c – a)[(a + b + c)(c – b)]

⇒ (a – b) (b – c) (c – a) (a + b + c)