$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real numbers such $ x+y+z = 0 $.
prove : $ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
Please think and write your solutions! ; )

If $x=y=z=0$, then it is trivial.
WLOG, $x\ne 0$ (at least one of variables  $\ne 0$).
Denote 

$y=ax$, 

$z=-x(1+a)$.
It is enough to prove, that
$$
\left(\frac{1^7+a^7-(1+a)^7}{7}\right)^2 = \left( \frac{1^5+a^5-(1+a)^5}{5} \right)^2 \cdot \left(\frac{1^4+a^4+(1+a)^4}{2}\right);
$$
$$
\left(\frac{7a+21a^2+35a^3+35a^4+21a^5+7a^6}{7}\right)^2 = \left( \frac{5a+10a^2+10a^3+5a^4}{5} \right)^2 \cdot \left(\frac{2+4a^2+6a^3+4a^4+2a^5}{2}\right);
$$
$$
(1+3a+5a^2+5a^3+3a^4+a^5)^2 = (1+2a+2a^2+a^3)^2 \cdot (1+2a+3a^2+2a^3+a^4);
$$
$$
(1+3a+5a^2+5a^3+3a^4+a^5) = (1+2a+2a^2+a^3) \cdot (1+a+a^2).
$$
Last identity is obvious.
