Find all solutions to sys. of equation

Question

I'm trying to find all solutions $x,y,z \in \mathbb{R}$ for the following system of equations

$x^3=y+y^3$ (1)

$y^3=z+z^3$ (2)

$z^3=x+x^3$ (3)

I found that $x+y+z=0$ by subtracting (2) and (3) from (1)

From the looks of it and by trial and error I suppose the answer is $x=0, y=0, z=0$

I have tried to play around with different substitutions and factoring different combinations of the three equations but to no success. If anyone could give me a tip or a solution to this problem that would be greatly appreciated! :)

Answer 1

I assume $x, y, z$ are reals.

Easy to see that $x,y,z$ must be all zero or have the same sign.

Suppose $x , y , z \gt 0$ solution. From (1) $x \gt y$ from (2) $y \gt z$ and from (3) $ z \gt x$ therefore $ x \gt x$ contradiction.

Similar for negatives.

Answer 2

solving the second equation for $y$ we get $$y=\sqrt[3]{z+z^3}$$ and $$z=\sqrt[3]{x+x^3}$$ and from here we get $$x^3=\sqrt[3]{\sqrt[3]{x+x^3}+x+x^3}+\sqrt[3]{x+x^3}+x+x^3$$ can you finish?