Solve the system of equations

$$
\begin{aligned}
& x^{3}=5 x+y \\
& y^{3}=5 y+x
\end{aligned}
$$
Solution. If $x=0$, then from the first equation $y=0$. Similarly, if $y=0$, then from the second equation $x=0$. Therefore, the pair $x_{1}=0, y_{1}=0$ is a solution to the system of equations, and we can assume from now on that $x \neq 0, y \neq 0$. Subtract the two equations from each other.

$$
\begin{gathered}
x^{3}-y^{3}=5 x+y-5 y-x \\
(x-y)\left(x^{2}+x y+y^{2}\right)=4(x-y)
\end{gathered}
$$

Now we need to consider two cases.

If $x=y$, then substituting into the original first equation

$$
x^{3}=5 x+x, \quad x^{3}-6 x=0
$$

Since we can assume that $x \neq 0$, we have $x^{2}=6$. From this, we get the solutions $x_{2}=y_{2}=\sqrt{6}$ and $x_{3}=y_{3}=-\sqrt{6}$.

If $x \neq y$, then $x^{2}+x y+y^{2}=4$. Before we proceed with a detailed examination of this, add the original two equations.

$$
\begin{aligned}
x^{3}+y^{3} & =6 x+6 y \\
(x+y)\left(x^{2}-x y+y^{2}\right) & =6(x+y)
\end{aligned}
$$

If now $x=-y$, then substituting into the second equation

$$
y^{3}=4 y, \quad y^{2}=4
$$

since the case $y=0$ has already been ruled out. From this, we get the solutions $x_{4}=2, y_{4}=-2$ and $x_{5}=-2, y_{5}=2$. We can assume from now on that $x+y \neq 0$, so the equation can be simplified by $(x+y)$. By subtracting and adding the equations, we have found four solutions and see that we only need to solve the following system of equations from now on:

$$
x^{2}+x y+y^{2}=4, \quad x^{2}-x y+y^{2}=6
$$

By subtracting the two equations, we get $x y=-1$. Adding this to the first equation and subtracting it from the second:

$$
x^{2}+2 x y+y^{2}=3, \quad x^{2}-2 x y+y^{2}=7
$$

From these two complete squares, we see that $x+y=\sqrt{3}, x+y=-\sqrt{3}, x-y=\sqrt{7}, x-y=-\sqrt{7}$. From the combinations of these, four additional systems of linear equations arise.

$$
\begin{array}{lllll}
x+y=\sqrt{3}, & x-y=\sqrt{7} & \Rightarrow & x_{6}=\frac{\sqrt{3}+\sqrt{7}}{2}, & y_{6}=\frac{\sqrt{3}-\sqrt{7}}{2} \\
x+y=\sqrt{3}, & x-y=-\sqrt{7} & \Rightarrow & x_{7}=\frac{\sqrt{3}-\sqrt{7}}{2}, & y_{7}=\frac{\sqrt{3}+\sqrt{7}}{2} \\
x+y=-\sqrt{3}, & x-y=\sqrt{7} & \Rightarrow & x_{8}=\frac{-\sqrt{3}+\sqrt{7}}{2}, & y_{8}=\frac{-\sqrt{3}-\sqrt{7}}{2} \\
x+y=-\sqrt{3}, & x-y=-\sqrt{7} & \Rightarrow & x_{9}=\frac{-\sqrt{3}-\sqrt{7}}{2}, & y_{9}=\frac{-\sqrt{3}+\sqrt{7}}{2}
\end{array}
$$

Upon verification, it can be seen that all nine pairs are indeed solutions to the system of equations.