Factoring a polynomial (multivariable) 
Factor $ (a - b)^3 + (b - c)^3 + (c-a)^3$ by SYMMETRY.

Okay, this is the problem. Let $f(a) = (a - b)^3 + (b-c)^3 + (c-a)^3$ obviously, if you let $a = b$ then, $f(b) = 0$, thus $(a - b)$ is a factor of $f(a)$. Then someone said :

If $(a - b)$ is a factor then $(b - c)$ and $(a-c)$ must be factors as well by symmetry.

But $f(a, b, c)$ is not symmetric, actually, $f(b, a, c) = -f(a, b, c)$ it is an alternating polynomial, so what is up with the solution?

The symmetry to use is not to change $f(a,b,c)$ to $f(b,a,c)$.
Instead, change $f(a,b,c)$ to $f(b,c,a)$ or $f(c,a,b)$.
This is the same as replacing $(a - b)^3 + (b - c)^3 + (c - a)^3$ by either 
$(b - c)^3 + (c - a)^3 + (a - b)^3$ or $(c - a)^3 + (a - b)^3 + (b - c)^3$,
which clearly makes no change in the value.
So the same reasoning that showed $(a-b)$ is a factor
must work for $(b-c)$ and $(c-a)$ as well.
And if $(c-a)$ is a factor, $(a - c)$ is a factor.
