<s>[INST] Finding all possible values of $x^4+y^4+z^4$ Given real numbers $x,y,z$ satisfying $x+y+z=0$ and 
$$ \frac{x^4}{2x^2+yz}+\frac{y^4}{2y^2+zx}+\frac{z^4}{2z^2+xy}=1$$
Find all possible values of $x^4+y^4+z^4$ with proof.
My attempt :
Putting $x=-y-z$ and doing subsequent calculations we get something like $$\displaystyle \sum_{cyc}\frac{x^4}{(z-x)(y-x)}=1$$
But nothing further. Can someone help me with this? Thanks in advance. 
 [/INST] this is a simple approach for $x^2+xy+y^2=1$, The op has a good start anyway, from his last step: 
note:$\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{y-z}\left(\dfrac{1}{z-x}-\dfrac{1}{y-x}\right)$
$\sum_{cyc}\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{(z-x)(y-z)}-\dfrac{x^4}{(y-z)(y-x)}+\dfrac{y^4}{(z-y)(x-y)}+\dfrac{z^4}{(x-z)(y-z)}=\dfrac{x^4-z^4}{(z-x)(y-z)}+\dfrac{y^4-x^4}{(y-z)(y-x)}=\dfrac{-(x+z)(x^2+z^2)+(y+x)(y^2+x^2)}{y-z}=\dfrac{y(x^2+z^2)-z(y^2+x^2)}{y-z}=\dfrac{x^2(y-z)-yz(y-z)}{y-z}=x^2-yz=x^2+xy+y^2$
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