Linear quadratic cubic system of equation

Question

I have been trying to make something out of this system of equations for quite sometimes now

$x+y+z=4\\x^2+y^2+z^2=38\\x^3+y^3+z^3=106$

I have tried direct substitution but the equation keeps expanding, and keeps getting complicated. I have also tried multipling by $x,$ $y,$ $z$ in several ways. But it was all a futile effort.

Any insight to this??

Answer 1

$$16=38+2(xy+xz+yz),$$ which gives $$xy+xz+yz=-11.$$ Thus, since $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz,$$ we obtain: $$106=64-3\cdot4\cdot(-11)+3xyz,$$ which gives $$xyz=-30$$ and $x$, $y$ and $z$ are roots of the equation: $$t^3-4t^2-11t+30=0.$$ Can you end it now?

I got $(x,y,z)=(2,-3,5)$ and all symmetric permutations of this.