<s>[INST] 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.
 [/INST] $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$
The key here is the identity (for all $n\in\mathbb Z_{\ge 3}$):
$$a^n+b^n+c^n=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-$$
$$-(ab+bc+ca)\left(a^{n-2}+b^{n-2}+c^{n-2}\right)+abc\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$$
Therefore: $$a^3+b^3+c^3=3abc\\ a^4+b^4+c^4=2(ab+bc+ca)^2\\ a^5+b^5+c^5=-5abc(ab+bc+ca)\\a^7+b^7+c^7=7abc(ab+bc+ca)^2$$
Therefore:
$$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2\\\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3}=\frac{a^5+b^5+c^5}{5}\\ \frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}\\$$
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