We can see that if $x=y$, then (1) is satisfied, and from (2),

$$
x=y=\sqrt[5]{105.5}
$$

is a solution to the system of equations. For the rest, we assume that $x \neq y$, so we can divide (1) by $(y-x)$:

$$
x y(x+y)\left(x^{2}+y^{2}\right)=-78
$$

Thus, neither $x$ nor $y$ can be zero, so we can talk about the number

$$
v=\frac{x}{y}+\frac{y}{x}
$$

Let's try to substitute this into our equations. From (1),

$$
-78=x^{2} y^{2}(x+y) \frac{x^{2}+y^{2}}{x y}=x^{2} y^{2}(x+y) v
$$

and from (2),

$$
\begin{gathered}
211=(x+y)\left(x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}\right)= \\
=x^{2} y^{2}(x+y)\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}+2-\frac{x}{y}-\frac{y}{x}-1\right)=x^{2} y^{2}(x+y)\left(v^{2}-v-1\right)
\end{gathered}
$$

By comparing these, we find that $v$ must satisfy the equation

$$
211 v=-78\left(v^{2}-v-1\right)
$$

Thus, the value of $v$ can only be $6 / 13$ or $-13 / 6$. Based on this, from (3), the possible values for $x / y$ are $-3 / 2$ and $-2 / 3$, which, according to (2), gives

$$
x=3, \quad y=-2 \quad \text { or } \quad x=-2, \quad y=3
$$

These values also satisfy (1), so together with $x=y=\sqrt[5]{211}$, they give all the solutions to the system of equations.
Solve the following system of equations:

$$
\begin{gathered}
y^{5} x-x^{5} y=78(x-y) \\
x^{5}+y^{5}=211
\end{gathered}
$$