<s>[INST] Linear quadratic cubic system of equation 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?? 
 [/INST] $$16=38+2(xy+xz+yz),$$ 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.
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