<s>[INST] 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?
 [/INST] 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.
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