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,
$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c) = 3 (x-a)(x-b)(x-c)$
$(x-a)^3 + (x-b)^3 + (x-c)^3 - 3 (x-a)(x-b)(x-c) = 0$. 
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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