$ \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! ; )

Let $x,y,z$ be the roots of $t^3-Qt - P=0$, $Q= -(xy+yz+zx), P = xyz$.
$0 = (x+y+z)^2 = -2Q + x^2+y^2+z^2 \implies x^2+y^2+z^2 = 2Q$.
You have $t^3 = P + Qt$ and when you replace with $x,y,z$ and sum:
$$
x^3+y^3+z^3 = 3P + Q(x+y+z) = 3P
$$
You also get the identities:
$$
x^4+y^4+z^4 = Q(x^2+y^2+z^2) = 2Q^2
\\
x^{n+3}+y^{n+3}+z^{n+3} = P(x^n+y^n+z^n) + Q(x^{n+1}+y^{n+1}+z^{n+1})
$$
Now return to the problem. Define $S_n = x^n+y^n+z^n$.
$$
LHS = \frac 1{49}S_7^2
= \frac 1{49}(PS_4+ QS_5)^2
= \frac 1{49}(2PQ^2 + Q(PS_2 + QS_3))^2\\
= \frac 1{49}(2PQ^2 + 2PQ^2 + 3PQ^2 )^2
= P^2Q^4
$$
$$
RHS = \frac1{50}S_5^2S_4
= \frac1{50}(PS_2+QS_3)^2 \times 2Q^2\\
= \frac1{50}(2PQ +3PQ)^2 \times 2Q^2
= P^2Q^4 = LHS
$$
