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*}

Here's your equations:
$x+y^2 = y^3$
$y+x^2 = x^3$
I could substitute
$x=y^3-y^2$
and get a degree 9 equation
in $y$,
but I'll try something else.
Looking at these,
I notice that if I subtract them,
I get something
in which everything
is divisible by
$x-y$.
Subtracting the second from the first,
I get
$(x-y)+(y^2-x^2)
=y^3-x^3
$
or
$(x-y)+(y-x)(y+x)
=(y-x)(y^2+xy+x^2)
$.
If $y \ne x$,
dividing by
$y-x$
gives
$-1+(y+x)
=y^2+xy+x^2
$
or
$0
=y^2+y(x-1)+x^2-x+1
$.
Solving this,
$\begin{array}\\
y
&=\dfrac{-(x+1)\pm \sqrt{(x-1)^2-4(x^2-x+1)}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{x^2-2x+1-4x^2+4x-4}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-3x^2+2x-3}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-2x^2-x^2+2x-1-2}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-2x^2-(x-1)^2-2}}{2}\\
\end{array}
$
Since the  discriminant is negative,
there are no real values of $y$.
Therefore the only real solution is
$x=y$.
