$12 \cdot 48$ Solve the equation
$$
4 x^{3}+4 x^{2} y-15 x y^{2}-18 y^{3}-12 x^{2}+6 x y+36 y^{2}+5 x-10 y=0
$$

for all positive integer solutions.
(30th International Mathematical Olympiad Shortlist, 1989)
［Solution］ Given the equation can be transformed into
$$
\begin{aligned}
& 4 x^{3}+4 x^{2} y-15 x y^{2}-18 y^{3}-12 x^{2}+6 x y+36 y^{2}+5 x-10 y \\
= & (x-2 y)\left(4 x^{2}+12 x y+9 y^{2}-12 x-18 y+5\right) \\
= & (x-2 y)(2 x+3 y-5)(2 x+3 y-1) \\
= & 0 .
\end{aligned}
$$

Then \(x-2 y=0\) or \(2 x+3 y-5=0\) or \(2 x+3 y-1=0\).
The equation
$$
x-2 y=0
$$

has positive integer solutions \((x, y)=(2 y, y), \quad y \in \mathbb{N}\).
The equation
$$
2 x+3 y-5=0
$$

has only one set of positive integer solutions
$$
(x, y)=(1,1) \text {. }
$$

The equation
$$
2 x+3 y-1=0
$$

has no positive integer solutions.
Therefore, all positive integer solutions to the given equation are
$$
\{(x, y)\}=\{(1,1)\} \cup\{(2 y, y), y \in \mathbb{N}\}
$$