**Zero**

If a+b+c=0 then a^3+b^3+c^3=3abc -----> (1)

Proof

a+b+c=0 => a+b=−c => (a+b)^3=(−c)^3 => a^3+b^3−3ab(a+b)=−c^3 => a^3+b^3−3ab(−c)=−c^3 => a^3+b^3+c^3=3abc

We have

x=a+b+c ---> (2)

(x-a)^3 + (x-b)^3 + (x-c)^3 - 3(x-a)(x-b)(x-c) ----> (3) =?

If you call x-a=a, x-b=b and x-c=c then (1)=(3) => x-a+x-b+x-c=0 => 3x=a+b+c ----> (4)

Comparing (2) and (4) x=3x => x=0 =>a+b+c=0 then as per (1) the answer in zero