How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks

Let $$f(x)=(x-b)^3 + (b-c)^3 + (c-x)^3 -3(x-b)(b-c)(c-x) $$
The $f$ is a polynomial of degree at most $3$. Moreover, $x^3$ appears twice, with coefficients $\pm 1$, thus cancel. It follows that $f$ is at most quadratic and it is obvious that
$$f(c)=f(b)=f(0)=0$$
Since $f$ is at most quadratic and has three roots, $f \equiv 0$.
Note: If you want to avoid the observation that the coefficients of $x^3$ cancel, then use $f(c)=f(b)=f(0)=0$ and
$$f(b+c)=c^3+(b-c)^3-b^3+3c(b-c)(-b)=c^3-b^3+(b-c)^3-3bc(b-c)$$
You can see that this is zero, or if you still want to avoid this computation, repeat the above argument:
Let $g(x)=x^3-b^3+(b-x)^3-3bx(b-x)$. Then $g$ is at most cubic and 
$$g(b)=g(0)=g(2b)=g(-b)=0 \Rightarrow g(x)\equiv 0$$
