A highschool factoring problem

Solution 1:

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.

Solution 2:

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.