How we can show this ;$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$ Let be $\quad x+y+z=0$
show this:
$$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$$
I solved ,but Im interesting what are you thinking about this,how can we arrive to solution quickly?
 A: Your polynomials are invariant by the action of $S_3$ on $(x,y,z)$.
A standard result says that the graded ring $\Bbb Q[x,y,z]^{S_3}$ is $\Bbb Q[x+y+z;xy+yz+xz;xyz]$.
Furthermore here, quotienting by $(x+y+z)$ we can work in the ring $(\Bbb Q[x,y,z]/(x+y+z))^{S_3} \cong \Bbb Q[xy+yz+xz;xyz]$.
Now in this graded ring, the pieces of degree $2,5,7$ are all of dimension $1$ (there is only one decomposition of $2,5,7$ as a sum of $2$s and $3$s).
Picking $x,y,z = 1,1,-2$ shows that $x^k+y^k+z^k$ are nonzero in that ring for $k > 1$, which proves that $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7) \pmod {x+y+z}$ for a rational number $q$ (and also that $(x^2+y^2+z^2)(x^3+y^3+z^3) = q' (x^5+y^5+z^5)$ but we don't particularly care).
Then you can easily check that when replacing $z$ with $-x-y$ in your expressions, the dominant coefficient in your polynomials is $1$ after doing the
divisions (note that for an odd prime $p$, by Fermat's little theorem, $(x^p+y^p+(-x-y)^p)/p$ even has integral coefficients, ... and the case $p=2$ just works), and you only have to check $1 \times 1 = 1$ to finish the proof.
You can also evaluate $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7)$ at $x,y,z = 1,1,-2$ to get $6\times (-30) = q \times (-126)$ hence $q=10/7$.
