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*}

The solution set of the second equation is the cubic parabola
$$\gamma:\quad y=x^3-x^2$$
which intersects the symmetry line $y=x$ in the three points
$$\left({1-\sqrt{5}\over2},{1-\sqrt{5}\over2}\right),\quad(0,0),\quad\left({1+\sqrt{5}\over2},{1+\sqrt{5}\over2}\right)\ .\tag{1}$$
Otherwise $\gamma$ lies in the interior of the four shaded regions of the following figure (note that for $x>0$ one has $x^3-x^2+x=x(x^2-x+1)>x(x-1)^2>0$, hence $x^3-x^2>-x$):

The solution set of the first equation is the curve $\hat\gamma$ obtained by reflecting $\gamma$ in the symmetry line $y=x$. It contains the three points $(1)$ and is otherwise a subset of the reflected shaded regions. Since the latter are disjoint from the original regions it follows that the intersection $\gamma\cap\hat\gamma$ is given by $(1)$.
