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 $T_{m}$ be $a^m+b^m+c^m$. 
Let $k=-ab-bc-ca$, and $l=abc$. 
Note that this implies $a,b,c$ are solutions to $x^3=kx+l$. 
Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)
It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$. 
From here, note that $T_{4}=2k^2$ using the identity above.  
In the same method, note that $T_{5}=5kl$.
From here, note $T_{7}=5k^2l+2k^2l=7k^2l$ from $T_{m+3}=kT_{m+1}+lT_{m}$ . Therefore, the equation simplifies to showing that $k^2l \times l=(kl)^2$, which is true.  
