<s>[INST] 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
 [/INST] We use the identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
let $x=a-b ,y=b-c, z=c-a$  we see $x+y+z=0$ so $(a-b)^3+(b-c)^3+(c-a)^3-3(a-b)(b-c)(c-a)=0 $
Done!
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