hello, could it be that your last equation of the given system must be $z^3=2x^3+x-2$? Sonnhard.

Yes, thanks Dr Sonhard. Edited.

Hint for solving the problem

Consider the function $f\left( t \right) = \sqrt[3]{{2{t^3} + t - 2}}$. The system of equations can be written in the form $\left\{ {\begin{array}{*{20}{c}} {x = f(y)} \\ {y = f(z)} \\ {z = f(x)} \\\end{array}} \right. \Rightarrow x = f\left( {f\left( {f\left( x \right)} \right)} \right)$ We have $f'\left( t \right) = \frac{{6{t^2} + 1}}{{3\sqrt[3]{{{{\left( {2{t^3} + t - 2} \right)}^2}}}}} > 0$ Assume that $f\left( x \right) > x$, we have $f\left( x \right) > x \Rightarrow f\left( {f\left( x \right)} \right) > f\left( x \right) > x \Rightarrow f\left( {f\left( {f\left( x \right)} \right)} \right) > f\left( x \right) \Rightarrow x > f\left( x \right)$ (which is not true) Assume that $f\left( x \right) < x$, in the same way we can prove that $f\left( x \right) > x$ (which is not true) So, we can conclude that $x=f(x)$ $ \Rightarrow {x^3} = 2{x^3} + x - 2 \Leftrightarrow {x^3} + x - 2 = 0 \Leftrightarrow x = 1$ Now, it's easy to get $x = y = z = 1$

we can solve it with no derivatives. if $y>1$,then $x^3-1=2(y^3-1)+(y-1)>2(y^3-1)$,so $x>1$,then analogously,wecan get $z^3-1>2(x^3-1)$,so$z>1$,hence $y^3-1>2(z^3-1)$,so $y^3-1>8(y^3-1)$,a contradiction! so $y\le 1$,similarly,$x\le 1,z\le 1$. on the other hand,adding the three equations yields $\sum x^3+\sum x=6$,so if one of the 3 $\le$ becomes $<$,then we can easily reach a contradiction.hence $x=y=z=1$.