# Can you prove this identity?

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?

-

It's simply $\ a^3\!+b^3\!-(a\!+\!b)^3=\,-3ab(a\!+\!b)\$ for $\ a,b\, =\, x\!-\!y,\ y\!-\!z$.

-

If one believes that the polynomial ring in 3 variables is a unique factorization ring, which it is, then we relatively easily see that both sides vanish when $x=y$ or when $x=z$ or when $y=z$. The constant $3$ can be determined by plugging in $x=0$, $y=1$, and $z=-1$, for example.

-
Oops. I didn't read this before I posted. (+1) – robjohn Feb 8 '14 at 6:17
+1 very pretty. – 6005 Feb 8 '14 at 18:17

$$a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - bc - ca - ab)$$

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 $a=b=c.$

-
Of course, elegance is subjective but I don't see this as elegant at all. Elegance is when you use simple concepts in a simple way to solve something that looks difficult; you're using (relatively) difficult concepts in a (relatively) difficult way to solve something simple. This seems to be way more work than just doing the obvious expansions. – David Richerby Feb 8 '14 at 17:23

$$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$$ $$=(a+b)^3+c^3-3ab(a+b)-3abc$$

$$=[(a+b)+c][(a+b)^2-(a+b)c+c^2]-3ab[(a+b)+c]$$

$$=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]$$

$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

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 $\displaystyle a-b=b-c=c-a=0\implies ?$

-