8. (2003 Nordic Mathematical Contest) Find all triples of integers $(x, y, z)$ such that $x^{3}+y^{3}+z^{3}-3 x y z=$ 2003.
8. Solution: The original equation can be transformed into
$$
\begin{array}{l}
(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4006=2 \times 2003, \\
\text { and }(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2),
\end{array}
$$

Therefore, $\left\{\begin{array}{l}x+y+z=1, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006 \text {; }\end{array}\right.$
or $\left\{\begin{array}{l}x+y+z=2003, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2 .\end{array}\right.$
For (1), we have $(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2}=4006$,
which simplifies to $6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006$.
However, $4006 \equiv 4(\bmod 6)$, which is a contradiction.
For (2), since $|x-y|,|y-z|,|z-x|$ include two 1s and one 0, without loss of generality, assume $x \geqslant y \geqslant z$.
When $x-1=y=z$, we have $3 y+1=2003$, which has no solution.
When $x=y=z+1$, we have $3 x-1=2003, x=668$.
Thus, the integer solutions that satisfy the conditions are $(668,668,667),(668,667,668),(667,668,668)$.