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.

Useful identities:
$(y-z)^{3}+(z-x)^{3}+(x-y)^{3}=
3(y-z)(z-x)(x-y)$
$(y-z)^{5}+(z-x)^{5}+(x-y)^{5}=
5(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)$
$(y-z)^{7}+(z-x)^{7}+(x-y)^{7}=
7(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)^{2}$
By letting $a=y-z,b=z-x,c=x-y$.
