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!

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 $x,y$
