Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:

If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$

There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.

Let $a,b,c$ be roots of $$x^3+mx-n=0$$
From Vieta formula we get:
$$ 0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$ 
so $$\boxed{a^2+b^2+c^2 =-2m}$$
Further $$x^3 = n-mx\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n $$
so $$\boxed{a^3+b^3+c^3 =3n}$$
Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so  $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$
so $$\boxed{a^5+b^5+c^5 =-5mn}$$
Finally $$x^7 = x(n-mx)^2 = n^2x-2mnx^2+m^2x^3 $$ $$=n^2x-2mnx^2+m^2n-m^3x$$
so $$\boxed{a^7+b^7+c^7 = 7m^2n}$$
and we are done.
