<s>[INST] Can you prove $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$? 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?
 [/INST] $$ a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - bc - ca - ab) $$
Note that this is the determinant of
$$ M \; = \;  
 \left(  \begin{array}{rrr}
  a  &  b  &  c \\
  c   &  a  &  b \\
  b  &  c   &  a   
\end{array} 
  \right),  
  $$
the matrix being evidently singular when $a=b=c$ or when $a+b+c=0.$
Meanwhile, with the evident possibility of permuting the letters, 
$$  (a^2 + b^2 + c^2 - bc - ca - ab)  = \frac{1}{4} \left( \, (a+b-2c)^2 + 3 (a-b)^2 \, \right)    $$
 and so is positive semidefinite only, it comes out to $0$ when $a=b=c.$
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