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:

$$(x-1)(y-1)(z-1)=x^2y^2z^2 &gt; 0$$

Here, there are two cases: all three variables are greater than $1$ or only one. 

If $x,y,z&gt;1$, it's easy to prove there are no solutions because $y^2&gt;x^3$ a.s.o. gives $xyz&lt;1$.

How can I prove there are no solutions when only variable, say $x$, is greater than $1$?
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.
Since only $x&gt;1$
Then $x&gt; y$

Using the first equation,

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

For the case $0\leq y &lt;1$, the RHS is negative,

So $y&lt;0$

Taking the third equation,

$z^3=x^2-x$

The RHS is always positive hence $0&lt;z&lt;1$

Taking the second equation,

$y^3=z^2-z$

Differentiate RHS to find stationary point,

$2z-1=0$

$z=1/2$, 

$-1/4\leq y^3 &lt;0$

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 $x&gt;1$