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!
 A: Hint
Using $z=-x-y$, the first equation write $2 x y+20=0$ and so $y=-\frac{10}{x}$. Now the second equation write, after minor simplifications, $$x^2+\frac{100}{x^2}=29$$ in which you recognize a quadratic in $x^2$.
I am sure that you can take from here.
A: 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 $y=-\frac{1}{120(z^2 + 8)}(120xz^2 + 960x + 3z^5 - 280z^3 - 1440z)$, $x^2 =z^2- y^2 +20$.
A: since$$x^3+y^3+z^3=3xyz\Longrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0$$
$$\Longrightarrow x+y+z=0 ,\text {or} ,x=y=z(\text{impossible})$$
let  $z=-x-y$,then you have
$$x^2+y^2-z^2=20\Longrightarrow x^2+y^2-(x+y)^2=20\Longrightarrow xy=-10$$
since
$$x^4+y^4=(x^2+y^2)^2-2x^2y^2=(20+z^2)^2-200=z^4+40z^2+200$$
so
$$z^4+40z^2+200-z^4=560\Longrightarrow z^2=9\Longrightarrow z=\pm 3$$
then It is easy to find $x,y$
