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

We know $$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$
If $x+y+z=0, x^3+y^3+z^3=-3(x+y)(y+z)(z+x)=-3(-z)(-x)(-y)=3xyz$
Alternatively if $x+y+z=0,$
$x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3=(-z)^3-3xy(-z)+z^3=3xyz$
Put $x=a-b,y=b-c,z=c-a$
