System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$ 
Solve over reals:
$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$

I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:
$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$
Here, there are two cases: all three variables are greater than $1$ or only one. 
If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.
How can I prove there are no solutions when only variable, say $x$, is greater than $1$?

Since only $x>1$
Then $x> y$
Using the first equation,
$x^3=y(y-1)$
For the case $0\leq y <1$, the RHS is negative,
So $y<0$
Taking the third equation,
$z^3=x^2-x$
The RHS is always positive hence $0<z<1$
Taking the second equation,
$y^3=z^2-z$
Differentiate RHS to find stationary point,
$2z-1=0$
$z=1/2$, 
$-1/4\leq y^3 <0$
Using the first equation again,
$x^3=y(y-1)$
Differentiating RHS to find min/max for $x$
$y=-1/4$,
$x^3=5/16$, which contradicts that $x>1$
