Task 3. Determine all pairs of positive integers $(x, y)$ for which

$$
x^{3}+y^{3}=4\left(x^{2} y+x y^{2}-5\right)
$$
Solution I. We can rewrite the equation as follows:

$$
(x+y)\left(x^{2}-x y+y^{2}\right)=4 x y(x+y)-20 .
$$

Now, $x+y$ is a divisor of the left side and of the first term on the right, so it is also a divisor of the second term on the right: $x+y \mid 20$. Since $x+y \geq 2$, this gives the possibilities for $x+y$ as $2,4,5,10,20$. If exactly one of $x$ and $y$ is even and the other is odd, the left side of the equation is odd and the right side is even, a contradiction. Therefore, $x+y$ is even, which rules out $x+y=5$. If $x+y=2$, then $x=y=1$, and the left side is positive while the right side is negative, so this possibility is also ruled out.

To try the other possibilities, we rewrite the equation a bit differently:

$$
(x+y)\left((x+y)^{2}-3 x y\right)=4 x y(x+y)-20 \text {. }
$$

If $x+y=4$, we have $4 \cdot(16-3 x y)=16 x y-20$, so $16-3 x y=4 x y-5$, thus $21=7 x y$, or $x y=3$. Therefore, $(x, y)=(3,1)$ or $(x, y)=(1,3)$. Both pairs satisfy the equation.

If $x+y=10$, we get $100-3 x y=4 x y-2$, so $7 x y=102$. But 102 is not divisible by 7, so this is not possible.

If $x+y=20$, we get $400-3 x y=4 x y-1$, so $7 x y=401$. But 401 is not divisible by 7, so this is not possible.

Having exhausted all possibilities, we conclude that $(1,3)$ and $(3,1)$ are the only solutions.