Example 4 Solve the system of equations $\left\{\begin{array}{l}x+y^{2}=y^{3}, \\ y+x^{2}=x^{3} .\end{array}\right.$
(57th Czech and Slovak Mathematical Olympiad)
【Analysis】Looking at each equation individually, they cannot be factored, but due to cyclic symmetry, the two equations can be factored after subtraction.
Solving the two equations by subtracting and rearranging, we get
$$
(x-y)\left(x^{2}+x y+y^{2}-x-y+1\right)=0 \text {. }
$$

Notice that,
$$
\begin{array}{l}
x^{2}+x y+y^{2}-x-y+1 \\
=\frac{1}{2}(x+y)^{2}+\frac{1}{2}(x-1)^{2}+\frac{1}{2}(y-1)^{2} \\
\geqslant 0,
\end{array}
$$

with equality holding if and only if $x+y=0$ and $x=y=1$. But this is impossible.
$$
\begin{array}{l}
\text { Hence } x^{2}+x y+y^{2}-x-y+1>0 \\
\Rightarrow x=y \Rightarrow x+x^{2}=x^{3} \\
\Rightarrow(x, y)=(0,0),\left(\frac{1 \pm \sqrt{5}}{2}, \frac{1 \pm \sqrt{5}}{2}\right) .
\end{array}
$$

Therefore, $(x, y)=(0,0),\left(\frac{1 \pm \sqrt{5}}{2}, \frac{1 \pm \sqrt{5}}{2}\right)$.