Example 4 Let real numbers $x, y, z$ simultaneously satisfy
$$
\left\{\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}\right.
$$

Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-2)^{2}$. ${ }^{[3]}$
(1st Youth Mathematical Week (Zonghu Cup) Mathematical Competition)
Solve: From the given, we have
$$
\left\{\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}\right.
$$

Multiplying the above three equations, we get
$$
(x-2)(y-2)(z-2)
$$
$$
\begin{aligned}
= & -6(x-2)(y-2)(z-2)(x+1)^{2}(y+1)^{2}(z+1)^{2} \\
\Rightarrow & (x-2)(y-2)(z-2) . \\
& {\left[1+6(x+1)^{2}(y+1)^{2}(z+1)^{2}\right]=0 } \\
\Rightarrow & (x-2)(y-2)(z-2)=0 .
\end{aligned}
$$

Without loss of generality, let $x-2=0$.
Substituting into the given equations, we find $y=z=2$.
Thus, $2008(x-1)^{2}+2009(y-1)^{2}+2010(z-2)^{2} = 4017$.