Example 1: Among the numbers $2008, 2009, 2010$, which can be written in the form $x^{3}+y^{3}+z^{3}-3 x y z$? (where $x, y, z$ are all positive integers)
Solve: Because $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z) \left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=(x+y+z) - \frac{1}{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$,

If $2008=x^{3}+y^{3}+z^{3}-3 x y z$, then $(x+y+z) \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4016 = 2^{4} \cdot 251$. Since $x+y+z$ and $\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$ are both integers, we can derive 10 systems of equations, of which only $\left\{\begin{array}{l}x+y+z=2008, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2,\end{array}\right.$ and $\left\{\begin{array}{l}x+y+z=502, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=8\end{array}\right.$ have positive integer solutions. These two systems of equations are symmetric in $x, y, z$. Without loss of generality, assume $x \leqslant y \leqslant z$, so we only write one set of solutions: $x=y=669, z=670$; $x=166, y=z=168$.

If $2009=x^{3}+y^{3}+z^{3}-3 x y z$, then $(x+y+z) \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4018 = 2 \cdot 7^{2} \cdot 41$. Since $x+y+z$ and $\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$ are both integers, we can derive 12 systems of equations, of which only $\left\{\begin{array}{l}x+y+z=2009, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2,\end{array}\right.$ and $\left\{\begin{array}{l}x+y+z=41, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=98\end{array}\right.$

have positive integer solutions. The two sets of solutions are $x=669, y=z=670$ and $x=9, y=z=16$.

It is known that 2010 is a multiple of 3.
If $3 \mid(x+y+z)$, then $3 \mid(x+y+z)^{2}$, so $3 \mid(x+y+z)^{2}-3(x y+y z+z x)$, i.e., $3 \mid\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$, hence $9 \mid x^{3}+y^{3}+z^{3}-3 x y z$, but 9 does not divide 2010.

If $3 \mid\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$, similarly, we can deduce that $9 \mid\left(x^{3}+y^{3}+z^{3}-3 x y z\right)$, but 9 does not divide 2010.

In summary, 2008 and 2009 can be expressed, but 2010 cannot be expressed.