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

Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-1)^{2}$.
4. Transform to get
$$
\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}
$$

Multiplying the above three equations yields
$$
\begin{array}{l}
(x-2)(y-2)(z-2) \\
=-6(x-2)(y-2)(z-2) . \\
(x+1)^{2}(y+1)^{2}(z+1)^{2} .
\end{array}
$$

It can only be that $(x-2)(y-2)(z-2)=0$.
Without loss of generality, let $x=2$. Then we get $x=y=z=2$.
Therefore, $2008(x-1)^{2}+2009(y-1)^{2}+2010(z-1)^{2}$ $=6027$.