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 $1$?
 A: If $xyz=0$ so  $(x,y,z)=(0,0,0).$
Let $xyz=r\neq0.$
Thus, by the  WE Tutorial School's work we obtain:
$$(x+1)(y+1)(z+1)=1.$$ or
$$x+y+z+xy+xz+yz=-r.$$
In another hand, by your work we obtain:
$$(x-1)(y-1)(z-1)=x^2y^2z^2,$$ which gives
$$x+y+z-xy-xz-yz=r^2-r+1,$$ which gives
$$x+y+z=\frac{(r-1)^2}{2}$$ and
$$xy+xz+yz=-\frac{r^2+1}{2}.$$
But after summing of our equations we obtain:
$$\sum_{cyc}(x^3-x^2+x)=0$$ and since
$$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$$ and $$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz),$$ we have: 
$$\frac{(r-1)^6}{8}+\frac{3(r-1)^2(r^2+1)}{4}+3r-\frac{(r-1)^4}{4}-r^2-1+\frac{(r-1)^2}{2}=0$$ or
$$r^6-6r^5+19r^4-24r^3+11r^2+6r+1=0$$ or
$$(r^3-3r^2+3r+1)^2+4r^2(r^2-2r+2)=0,$$ which is impossible for real value of $r$.
Id est, $(0,0,0)$ is an unique solution of the system.
A: I will prove by contradiction that there exists no solutions for $x > 1$ and $y,z < 1$. Suppose $(x,y,z)$ is such a solution.
Iteration #1: First, note that:
$$
z^3 = x^2 - x = x(x - 1) > 0 \implies z > 0
$$
Thus, $0 < z < 1$. We see that:
$$
y^3 = z^2 - z = z(z - 1)
$$
For $0 < z < 1$, $-\frac{1}{4} < z(z - 1) < 0$, so $-2^{-\frac{2}{3}} < y < 0$. For this range of values of $y$, since $y(y - 1)$ is decreasing for $y < 0$, we have that:
$$
x^3 = y^2 - y < 2^{-\frac{4}{3}} + 2^{-\frac{2}{3}} = 1.0268108 \implies x < 1.00885823
$$
Iteration #2: Applying this inequality to the third equation again, we have:
$$
z^3 = x(x - 1) < 1.00885823(1.00885823 - 1) = 0.0089367 \implies z < 0.20752
$$
For $0 < z < 0.20752$, we have that:
$$
0.20752(0.20752 - 1)  < y^3 < 0 \implies -0.54787 < y < 0
$$
Once again, we have:
$$
x^3 = y(y - 1) < -0.54787 \left(-0.54787 - 1\right) < 1 \implies x < 1
$$
This contradicts that $x > 1$. Therefore, no non-trivial solutions exist.
A: 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 $x>1$
