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*}

Since $y=x^3-x^2$, we get $x+(x^3-x^2)^2=(x^3-x^2)^3$ or
$$x(x^2-x-1)(x^6-2x^5+2x^4-2x^3+2x^2-x+1)=0$$
and since $x^6-2x^5+2x^4-2x^3+2x^2-x+1=(x^6-2x^5+x^4)+(x^4-2x^3+x^2)+(x^2-x+1)>0$,
we get your answer.
