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.

Start off with the fact that 
$$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca)
$$
Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term:
$$
x^3+kx+m = 0
$$ with $ab+bc+ca = k$. And by the way, that says that 
$$a^2+b^2+c^2=-2k$$
Now start chaining upward, expressing $a^n+b^n+c^n$ in terms of $m$ and $k$.  For $n=3$ we add the cubic expression with $x=a$ to that with $x=b$ and $x=c$ to get
$$
a^3+ b^3 + c^3 +k(a+b+c)+(m+m+m) = 0 \\
a^3+ b^3 + c^3 +3m = 0 \\
a^3+ b^3 + c^3 = -3m 
$$
Now for $n=4$ we multiply each of the equations by $a$, $b$ and $c$ respectively before adding them:
$$
a^4+ b^4 + c^4 +k(a^2+b^2+c^2)+m(a+b+c) = 0 \\
a^4+ b^4 + c^4 -2k^2 = 0 \\
a^4+ b^4 + c^4 = 2k^2
$$
Now for $n=5$ we multiply each of the equations by $a^2$, $b^2$ and $c^2$ respectively before adding them:
$$
a^5+ b^5 + c^5 +k(a^3+b^3+c^3)+m(a^2+b^2+c^2) = 0 \\
a^5+ b^5 + c^5 -3mk  -2mk = 0 \\
a^5+ b^5 + c^5 = 5mk  
$$
For this problem we can afford to skip $n=6$.
$$
a^7+ b^7 + c^7 +k(a^5+b^5+c^5)+m(a^4+b^4+c^4) = 0 \\
a^7+ b^7 + c^7 +5mk^2 +2mk^2 = 0 \\
a^7+ b^7 + c^7 = -7mk^2   
$$
Then your identity reads $$
\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\
\left( -m \right) \left( -mk^2 \right) = (mk)^2 $$
