Solve the following system of equations:

$$
\begin{aligned}
x-\frac{1}{(x-y)^{2}}+y & =-10 \\
x y & =20 .
\end{aligned}
$$
First, let's pay attention to the fact that the roles of $x$ and $y$ can be interchanged. This is obvious from the second equation, but it is hidden in the first one: $x+y$ includes a fraction whose value does not change when $x$ and $y$ are swapped. From this realization, the idea can arise to introduce the notation $u=(x-y)^{2}$. Starting from the identity $(x+y)^{2}=(x-y)^{2}+4 x y$, and substituting the value of $x y$ from the second equation, we get the equation

$$
(x+y)^{2}=u+80
$$

From the first equation of the system of equations:

$$
(x+y)^{2}=\left(\frac{1}{u}-10\right)^{2}
$$

Equating the two expressions, we get

$$
u+80=\left(\frac{1}{u}-10\right)^{2}
$$

Rearranging the equation, we get

$$
u^{3}-20 u^{2}+20 u-1=0
$$

This equation is a so-called reciprocal equation, so if $u$ is a solution, then $\frac{1}{u}$ is also a solution.

Based on the known identity $u^{3}-1=(u-1)\left(u^{2}+u+1\right)$, it is easy to factorize the above equation:

$$
(u-1)\left(u^{2}-19 u+1\right)=0
$$

The solutions of the equation are: $u_{1}=1, u_{2}=\frac{19+\sqrt{357}}{2} \approx 18.9472, u_{3}=\frac{19-\sqrt{357}}{2} \approx 0.0528$.

Due to the above, the resulting quadratic equation will also be a reciprocal equation, which is why its solutions are reciprocals of each other.

Substituting the identities $x=\frac{x+y+x-y}{2}$ and $y=\frac{x+y-(x-y)}{2}$ with the relationships $(x-y)^{2}=u, x+y=\frac{1}{u}-10$, we get

$$
x=\frac{\frac{1}{u}-10 \pm \sqrt{u}}{2}, \quad y=\frac{\frac{1}{u}-10 \mp \sqrt{u}}{2}
$$

Since all three values of $u$ are positive, substituting each of them into the (total of 6) number pairs, both equations hold. Consider any arbitrary $(x, y)$ pair among them, for which $x+y=\frac{1}{u}-10, (x-y)^{2}=u$. Substituting into the first equation,

$$
x-\frac{1}{(x-y)^{2}}+y=\frac{1}{u}-10-\frac{1}{u}=-10
$$

Furthermore,

$$
x y=\frac{(x+y)^{2}-(x-y)^{2}}{4}=\frac{\left(\frac{1}{u}-10\right)^{2}-u}{4}=\frac{\frac{1}{u^{2}}+100-\frac{20}{u}-u}{4}
$$

From the equation $(*)$ satisfied by $u$, it is clear that

$$
\frac{1}{u^{2}}+100-\frac{20}{u}-u=80
$$

so indeed $x y=20$.

Thus, the solutions of the system of equations are: $\mathrm{x}_{1}=-4 \quad y_{1}=-5 \quad x_{3} \approx-2.7972 \quad y_{3} \approx-7.15 \quad x_{5} \approx 4.5884 \quad y_{5} \approx 4.3588$ $x_{2}=-5 \quad y_{2}=-4 \quad x_{4} \approx-7.15 \quad y_{4} \approx-2.7972 \quad x_{6} \approx 4.3588 \quad y_{6} \approx 4.5884$.