If `3x=a+b+c ,` then the value of `(x-a)^3+\ (x-b)^3+\ (x-c)^3-3\ (x-a)(x-b)(x-c)` is (a)`a+b+c` (b) `(a-b)(b-c)(c-a)` (c)`0` (d) None of these

Explanation

The given expression is
`(x − a)^3 + (x −b)^3 + (x − c)^3− 3(x − a) (x − b) (x −c)`
Recall the formula:
`a^3+b^3+c^3−3abc=(a+b+c)(a^2+b^2+c^2−ab−bc+ca)`
Using the above formula the given expression becomes
`(x − a)^3 + (x −b)^3 + (x − c)^3 − 3(x − a) (x − b) (x −c)`
`{(x−a)+(x−b)+(x−c)}{(x−a)^2+(x−b)^2+(x−c)^2−(x−a)(x−b)−(x−b)(x−c)−(x−c)(x−a)}`
`(x−a+x−b+x−c){(x−a)^2+(x−b)^2+(x−c)^2−(x−a)(x−b)−(x−b)(x−c)−(x−c)(x−a)}`
Given that
`3x=(a+b+c)`
`⇒3x−a−b−c=0`
Therefore the value of the given expression is
`(x − a)^3 + (x −b)^3 + (x − c)^3 − 3(x − a) (x − b) (x −c)`
`
=0.{(x−a)^2+(x−b)^2+(x−c)^2−(x−a)(x−b)−(x−b)(x−c)−(x−c)(x−a)}`
`=0`
Therefore, option C is correct.