Find all solutions of the system of equations: $(x+y)^{3}=z,(y+z)^{3}=x,(z+x)^{3}=y$.

#
Subtracting the second equation from the first, we get: $(x-z)\left((x+y)^{2}+(x+y)(y+z)+(y+z)^{2}\right)=-(x-z)$.

Since an expression of the form $a^{2}+a b+b^{2}$ cannot be negative, $x=z$. Similarly, $x=y$.

It remains to solve the equation $(2 x)^{3}=x$.

## Answer

$(0,0,0),\left( \pm \frac{\sqrt{2}}{4}, \pm \frac{\sqrt{2}}{4}, \pm \frac{\sqrt{2}}{4}\right)$.