1.

The value of \(\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \) is :(a) 1 (b) 3 (c) 1/3 (d) Zero

Answer»

(b) 3

Since a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc. 

So, as (a – b) + (b – c) + (c – a) = 0 

⇒ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)

∴ Given expression = \(\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\) = 3.


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