<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] Let $ a-b = x $ , $ b -c= y $ and $  c-a =z$ . So actually you want to know the value of  $ x^3+y^3+z^3-3xyz$ . Here the technique resides . 
$ x+y = (a-c) =-z $ ,
$x^3+y^3+z^3-3xyz = (x+y)^3-3xy(x+y) +z^3-3xyz
                   = -z^3 +z^3 +3xyz -3xyz  = 0 $
Hence the value is $ 0 $
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