TITLE: Can you prove $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$?
QUESTION [10 upvotes]: Show that $$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$$
This can be shown through expansion but there is a more elegant solution
I cannot discover anything I would consider elegant. Can anyone help?

REPLY [2 votes]: $$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$$
$$=(a+b)^3+c^3-3ab(a+b)-3abc$$
$$=[(a+b)+c][(a+b)^2-(a+b)c+c^2]-3ab[(a+b)+c]$$
$$=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
Though not required in the current case, $$a^2+b^2+c^2-ab-bc-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}2$$ which will be zero iff $\displaystyle a-b=b-c=c-a=0\implies ?$