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

Notice that this is invarient if the same constant is added to all three variables, so one of them may be set equal to zero. Set $c=0$. Then the expression becomes $(a-b)^3 -a^3 +b^3+3ab(a-b)$, easily seen to be zero.
   BTW, apparently the original expression is equivalent to saying $x^3 +y^3+z^3 =3xyz$ if $x+y+z=0$.
EDIT:Yes, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$
