Solve in real numbers the system

$\left\{\begin{array}{l}\mathrm{x}+{\mathrm{y}}^2={\mathrm{y}}^3\\ \mathrm{y}+{\mathrm{x}}^2={\mathrm{x}}^3\end{array}\right.$.

Then the number of real solutions are ________ 

We will certainly not try to replace $\mathrm{x}={\mathrm{y}}^3-{\mathrm{y}}^2$ in the second equation, since that would give a ninth degree equation! Instead, let us take the difference of the two equations and factor x-y. We obtain

$\mathrm{x}-\mathrm{y}-\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)=-\left({\mathrm{x}}^3-{\mathrm{y}}^3\right)$

or equivalently

$(\mathrm{x}-\mathrm{y})(1-\mathrm{x}-\mathrm{y})=-(\mathrm{x}-\mathrm{y})\left({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2\right)$

Suppose first that $\mathrm{x}-\mathrm{y}=0$. Since $\mathrm{x}=\mathrm{y}$, the system reduces to $\mathrm{x}+{\mathrm{x}}^2={\mathrm{x}}^3$. This equation has the obvious solution $\mathrm{x}+{\mathrm{x}}^2={\mathrm{x}}^3$ and after
factoring out x we obtain the equation $1+\mathrm{x}={\mathrm{x}}^2$, with solutions $\frac{1\pm \sqrt{5}}{2}$. This gives us three solutions of the system.