<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] 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.
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