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?

Here is one way of making progress, which uses the cubic as part of the solution. There are other routes which involve knowing some standard factorisations.
First note that $z=-(x+y)$ from the first equation and substitute in the second, obtaining:
$$-3x^2y-3xy^2 = 9$$
Divide by 3 to get:
$$-xy(x+y) = xyz = 3$$
Now $x,y,z$ are the roots of the cubic equation $t^3+kt-3 = 0$, 
and therefore satisfy $t^5+kt^3-3t^2=0$
Substitute $x,y,z$ successively into this equation and add to get
$$30+9k-3(x^2+y^2+z^2) = 0$$
And use $0=(x+y+z)^2=x^2+y^2+z^2+2k$ to finish.
