12.87 Determine all integer solutions of the indeterminate equation
$$
x^{3}+x^{2} y+x y^{2}+y^{3}=8\left(x^{2}+x y+y^{2}+1\right)
$$

(Luxembourg and four other countries' International Mathematical Competition, 1980)
【Solution】Given the equation can be transformed into
$$
\left(x^{2}+y^{2}\right)(x+y-8)=8 x y+8 \text {. }
$$

If $x$ and $y$ are one odd and one even, then the left side of (1) is odd, and the right side is even, which is impossible.
Therefore, $x$ and $y$ have the same parity.
Thus, $x+y-8$ is even.
(1) If $x+y-8 \geqslant 6$, then
$$
\begin{aligned}
x^{2}+y^{2} \geqslant \frac{(x+y)^{2}}{2} & \geqslant \frac{14^{2}}{2}>4, \\
\left(x^{2}+y^{2}\right)(x+y-8) & \geqslant 6\left(x^{2}+y^{2}\right) \\
& \geqslant 2\left(x^{2}+y^{2}\right)+8 x y \\
& >8+8 x y .
\end{aligned}
$$

In this case, equation (1) does not hold, and the equation has no integer solutions.
(2) If $x+y-8 \leqslant-4$, then
$$
\left(x^{2}+y^{2}\right)(x+y-8) \leqslant-4\left(x^{2}+y^{2}\right) \leqslant 8 x y<8 x y+8 \text {. }
$$

Similarly, equation (1) does not hold, and the equation has no integer solutions.
(3) If $x+y-8=4$, then from (1) we get
$$
(x-y)^{2}=2 \text {. }
$$

The equation has no integer solutions.
(4) If $x+y-8=2$, then from (1) we get
$$
x^{2}+y^{2}=4 x y+4,
$$

Thus, we solve
$$
\left\{\begin{array} { l } 
{ x = 8 , } \\
{ y = 2 , }
\end{array} \quad \left\{\begin{array}{l}
x=2 \\
y=8
\end{array}\right.\right.
$$
(5) If $x+y-8=0$, then we have
$$
8 x y+8=0 .
$$

Clearly, there are no integer solutions.
(6) If $x+y-8=-2$, then from (1) we get
$$
x^{2}+y^{2}+4 x y+4=0 \text {. }
$$

Thus, we have
$$
x+y=6, \quad x y=-20 \text {. }
$$

There are still no integer solutions.
Therefore, the given equation has only two sets of integer solutions:
$$
(x, y)=(8,2) \text { or }(2,8) \text {. }
$$