<s>[INST] I have a Hard time solving this system of nonlinear equations $x^2+y^2-z^2=20$
$x^4+y^4-z^4=560$
$x^3+y^3+z^3=3xyz$
I know the fact that if $x^3+y^3+z^3=3xyz$ then $x+y+z=0$ (coming from Euler's identity) and first equation can be written as 
$(x+y-z)^2-2(xy-xz-yz)=20$ and since $x+y=-z$ then $(xy-xz-yz)=2z^2-10$. 
And the second equation becomes 
$(x^2+y^2-z^2)^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$. 
Since we know $x^2+y^2-z^2=20$, therefore $20^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$ or 
$(x^2y^2-2x^2z^2-2y^2z^2)=80$. 
Now I am stuck! Help please!
 [/INST] This is a well-knonw example for solving a system of polynomial equations via Groebner bases. The real solutions are given by (check this yourself):
$$
(x,y,z)=(5,-2,-3),\; (-2,5,-3),\;(2,-5,3),\;(-5,2,3).
$$
Over the complex numbers, there are another $8$ solutions for $z$, i.e., the roots of the polynomial
$$
9z^8 - 240z^6 + 1600z^4 + 57600z^2 + 230400=0,
$$
and $y=-\frac{1}{120(z^2 + 8)}(120xz^2 + 960x + 3z^5 - 280z^3 - 1440z)$, $x^2 =z^2- y^2 +20$.
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