Task 3. Determine all possible values of $\frac{1}{x}+\frac{1}{y}$ if $x$ and $y$ are real numbers (not equal to 0) that satisfy $x^{3}+y^{3}+3 x^{2} y^{2}=x^{3} y^{3}$.
Solution. We rewrite the given to $x^{3}+y^{3}-x^{3} y^{3}=-3 x^{2} y^{2}$, from which we find

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

By using that $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, we can also write

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

If we now equate the two right-hand sides, we see that either $x+y-x y=0$ or $x^{2}+y^{2}+2 x y+x^{2} y+x y^{2}+x^{2} y^{2}=3 x y$. The latter case can be rewritten as

$$
\begin{aligned}
0 & =x^{2}+y^{2}-x y+x^{2} y+x y^{2}+x^{2} y^{2} \\
& =\frac{1}{2}(x-y)^{2}+\frac{1}{2} x^{2}\left(y^{2}+2 y+1\right)+\frac{1}{2} y^{2}\left(x^{2}+2 x+1\right) \\
& =\frac{1}{2}(x-y)^{2}+\frac{1}{2} x^{2}(y+1)^{2}+\frac{1}{2} y^{2}(x+1)^{2} .
\end{aligned}
$$

The sum of three squares must therefore be 0, which implies that each of the squares must be 0. Thus, $x=y=-1$. In this case, the given equation is indeed satisfied and $\frac{1}{x}+\frac{1}{y}=-2$.

In the other case, $x+y-x y=0$, so $\frac{1}{y}+\frac{1}{x}-1=0$, which implies that $\frac{1}{x}+\frac{1}{y}=1$. This is satisfied, for example, for $x=y=2$.

We conclude that the desired values are -2 and 1.

Alternative approach. It is also possible to find the desired form of the given relation with less algebraic manipulation. We describe a possible approach here.

Dividing by $x^{3} y^{3}$ gives

$$
\frac{1}{x^{3}}+\frac{1}{y^{3}}+\frac{3}{x y}=1
$$

Here we already see something with $\frac{1}{x}$ and $\frac{1}{y}$ in it. We substitute $a=\frac{1}{x}$ and $b=\frac{1}{y}$ to make the notation easier. Then we have

$$
a^{3}+b^{3}+3 a b=1
$$

and we are looking for the value of $a+b$. There are probably solutions with $a=b$. We get $2 a^{3}+3 a^{2}=1$. Trying numbers or factoring this gives that $a=-1$ and $a=\frac{1}{2}$ satisfy it. Now $a+b$ is equal to -2 or 1. We can now guess that perhaps all pairs $(a, b)$ with $a+b=-2$ or all pairs $(a, b)$ with $a+b=1$ work; then the expression would be divisible by $a+b+2$ or $a+b-1$, respectively. Trying shows that the latter is indeed the case. We can rewrite it as

$$
(a+b-1)\left(a^{2}+b^{2}-a b+a+b+1\right)=0 .
$$

We also know by now that $a+b=-2$ does not give a solution for all $a$ and $b$, but does for $a=b=-1$. So we would want something with $a+1$ and $b+1$ in the right factor. It is now easy to see that the right factor is the sum of three squares. We can now finish it (similar to above).