A system of polynomial equations with odd degree

Question

Consider real numbers $u_1, u_3,u_5 \in \mathbb{R}$ and the polynomial system

$ x + y +z= u_1 \\ x^3 + y^3 +z^3= u_3 \\ x^5 + y^5 +z^5= u_5 \\ $

How to prove that the polynomial system has only one solution up to permutations?

Thanks in advance.

Answer 1

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.

Hence, $x^3+y^3+z^3=27u^3-27uv^2+3w^3$ and

$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3$

Thus, we get an easy system on $v^2$ and $w^3$ and the rest is proving of a cubic equation.