$ \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! ; )
 A: 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.
A: 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 $S_n = x^n+y^n+z^n$.
$$
LHS = \frac 1{49}S_7^2
= \frac 1{49}(PS_4+ QS_5)^2
= \frac 1{49}(2PQ^2 + Q(PS_2 + QS_3))^2\\
= \frac 1{49}(2PQ^2 + 2PQ^2 + 3PQ^2 )^2
= P^2Q^4
$$
$$
RHS = \frac1{50}S_5^2S_4
= \frac1{50}(PS_2+QS_3)^2 \times 2Q^2\\
= \frac1{50}(2PQ +3PQ)^2 \times 2Q^2
= P^2Q^4 = LHS
$$
A: HINT:
Let $x,y,z$ be the roots of $\displaystyle  t^3+bt+c=0\ \ \ \ (1)$
$\displaystyle\implies xy+yz+zx=b, xyz=-c$
Multiplying $(1)$ by $t^n\ne0$
$\displaystyle \implies t^{n+3}+bt^{n+1}+ct^n=0$
$\displaystyle \implies\sum x^{n+3}=-b\sum x^{n+1}-c\sum x^n$
$\displaystyle n=1\implies \sum x^4=-b\sum x^2-c\sum x =-b[(\sum x)^2-2(xy+yz+zx)]=-b(-2b)=2b^2$
From $\displaystyle n=0\implies \sum x^3=-b\sum x-3c=-3c$
$\displaystyle n=2\implies \sum x^5=-b\sum x^3-c\sum x^2 =-b(-3c)-c[(\sum x)^2-2(xy+yz+zx)]=3bc-c(-2b)=5bc$
Can you take home from here? 
