# Find the value of $(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)$ if $a+b+c = 3x$

Given :

$a + b + c = 3x$

To find :

We have to find the value of $(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)$.

Solution :

We know that,

If $a + b + c = 0$ then $a^3+ b^3+ c^3=3abc$.

$(x-a)+(x-b)+(x-c)= 3x-(a+b+c)$

$= 3x-3x$

= 0

Therefore,

$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c) = 3 (x-a)(x-b)(x-c)$

$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c) = 0$.â€Š

The value of $(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c)$ is 0.
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