10. (40 points) Try to determine: Among the numbers $2008$, $2009$, and $2010$, which can be expressed in the form $x^{3}+y^{3}+z^{3}-3 x y z$ $\left(x, y, z \in \mathbf{N}_{+}\right)$?
10. When $x=670, y=z=669$, we get 2008; when $x=669, y=z=670$, we get 2009.
2010 cannot be expressed in this form.
The proof below uses proof by contradiction:
If $31 t$ and $9 \times t$, then $t$ cannot be written in the form $x^{3}+y^{3}+z^{3}-3 x y z$.
If not, then because $3 \mid t$, we have
$$
31(x+y+z)
$$

or $3 \mid\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$.
But $3|(x+y+z) \Leftrightarrow 3|(x+y+z)^{2}$
$\Leftrightarrow 3 \mid\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$.
Thus, either form can lead to $9 \mid t$, which is a contradiction.
In summary, 2008 and 2009 can be expressed, but 2010 cannot.
Note: In fact,
$$
\begin{array}{l}
(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
=x^{3}+y^{3}+z^{3}-3 x y z .
\end{array}
$$

Let $x=a+1, y=z=a$, then the above expression is $3 a+1$;
Let $x=a-1, y=z=a$, then the above expression is $3 a-1$;
Let $x=a+1, y=a, z=a-1$, then the above expression is
$9 a$.
Thus, numbers of the form $3 a+1$, $3 a-1$, and $9 a$ can all be expressed.