System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
  x+y^2 &= y^3\\
  y+x^2 &= x^3
\end{align*}
These are the solutions:
 \begin{align*}
  (0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2); 
\end{align*}

Notice that $x=0 \implies y=0$, hence $(0,0)$ is a trivial solution. We will consider the case where $x \neq 0$. 
$$x=y^2(y-1)$$
$$y=x^2(x-1)$$
Suppose on the contrary that $x=1$, then from the second equation $y=0$ which contradicts the first equation $x=0$. Hence $x \neq 1$.
Suppose on the contrary that $x \in (0,1)$, 
$$y=x^2(x-1) \in (-1,0)$$
and hence $$x=y^2(y-1)<0$$
which is a contradiction.  Hence $x \notin (0,1).$ By symmetry, $y \notin (0,1)$.
Also, notice if $x>1$, then $y=x^2(x-1)>0$, and hence $y>1$.
and if $x<0$ then $y=x^2(x-1)<0$.
$$x=y^2(y-1)$$
$$x^2(x-1)=y$$
Multiply the two equations together, we have
$$x^3(x-1)=y^3(y-1)$$
Consider the function, $$f: (1,\infty) \rightarrow (0,\infty),\text{ where }f(t)= t^3(t-1).$$
We can easily see that this function is increasing and it is an injective function.
Hence, if $x>1$, we have $f(x)=f(y)$ and hence $x=y$.
Similarly, we can consider the function, $$g: (-\infty,0) \rightarrow (0,\infty),\text{ where }g(t)= t^3(t-1).$$
We can easily see that this function is decreasing and it is an injective function.
Hence, if $x<0$, we have $g(x)=g(y)$ and hence $x=y$.
Hence, we always have $x=y$.
$$x+x^2=x^3$$
$$x(1-x-x^2)=0$$
$x=0$ or $1-x-x^2=0$
