Find all integer solution pairs $(x, y)$ for $y^{3}=8 x^{6}+2 x^{3} y-y^{2}$ when $x \in$ $[0,10]$.
Solve: From $y^{3}=8 x^{6}+2 x^{3} y-y^{2}$
$\Rightarrow 8 y^{3}=64 x^{6}+16 x^{3} y-8 y^{2}$
$\Rightarrow y^{2}(8 y+9)=\left(8 x^{3}+y\right)^{2}$
$\Rightarrow 8 y+9=\left(\frac{8 x^{3}}{y}+1\right)^{2}$.
From equation (1), we know $y \mid 8 x^{3}$, and $\frac{8 x^{3}}{y}$ is even.
Thus, $\left\{\begin{array}{l}8 x^{3}=2 k y, \\ 8 y+9=(2 k+1)^{2}\end{array} \quad(k \in \mathbf{Z})\right.$
$\Rightarrow\left\{\begin{array}{l}y=\frac{k^{2}+k-2}{2}, \\ (2 x)^{3}=k^{3}+k^{2}-2 k .\end{array}\right.$
When $k \geq 2$, we have $k^{3}<k^{3}+k^{2}-2 k<(k+1)^{3}$.
Therefore, equation (2) has no integer solutions.
Thus, $k \in\{-2,-1,0,1,2\}$.
Substituting each value for inspection, we find:
When $k=-2,1$, $(x, y)=(0,0)$;
When $k=-1$, equation (2) has no integer solutions;
When $k=0$, $(x, y)=(0,-1)$;
When $k=2$, $(x, y)=(1,2)$.
In summary, the solutions are
$$
(x, y)=(0,0),(0,-1),(1,2) .
$$