A highschool factoring problem $x+y+z=0$
$x^3+y^3+z^3=9$
$x^5+y^5+z^5=30$
$xy+yz+zx=?$
I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?

We have the Newton-Girard identities $$x^3+y^3+z^3=(x+y+z)^3+3xyz-3(x+y+z)(xy+xz+yz)$$ and $$\begin{split}x^5+y^5+z^5=&(x+y+z)^5-5(x+y+z)^3 (xy+xz+yz)+\\5(x+y+z)&(xy+xz+yz)^2-5xyz(xy+xz+yz)+5xyz(x+y+z)^2\end{split}$$
Replacing all instances of $x+y+z$ with $0$, we have the simultaneous equations
$$\begin{align*}
3xyz&=9\\
-5xyz(xy+xz+yz)&=30
\end{align*}$$
You should now be able to solve for what you need.
