The value of \(\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}\) is(

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) \(\frac13\)(d) zero
Answer» (b) 3 Since (a-b) + (b-c) + (c-a) = 0 ∴ (a-b)³ + (b-c)³ + (c-a)³= 3(a-b)(b-c)(c-a) ⇒ \(\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}\) = \(\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\) = 3.

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) \(\frac13\)(d) zero

Answer»

(b) 3

Since (a-b) + (b-c) + (c-a) = 0

∴ (a-b)³ + (b-c)³ + (c-a)³= 3(a-b)(b-c)(c-a)

⇒ \(\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}\) = \(\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\) = 3.