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)