<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] If we remember the well-known formulas
$(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$
$(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy + y^2)$
$(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$
then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides.  (Under the change of notation $a,b,c \to -(x+y),x,y$)
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