3. Given the polynomial
$$
\begin{aligned}
a_{0}+ & \left(a_{1}+4\right) x+ \\
& \left(a_{2}-10\right) x^{2}+\left(a_{3}+6\right) x^{3}+\left(a_{4}-1\right) x^{4}+ \\
& \left(a_{5}-1\right) x^{5}+a_{6} x^{6}+\cdots+a_{2 \alpha 5} x^{2 \omega 5}
\end{aligned}
$$

can be divided by $x^{2}+3 x-2$, and $\alpha^{2}+3 \alpha-2=0$. Then
$$
a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2 \alpha 6} \alpha^{20 \% 5}
$$

has the value $\qquad$ .
3.0 .

Since $\alpha^{2}+3 \alpha-2=0$, $\alpha$ is a root of the equation $x^{2}+3 x-2=0$. Therefore,
$$
\begin{array}{l}
a_{0}+\left(a_{1}+4\right) \alpha+\left(a_{2}-10\right) \alpha^{2}+\left(a_{3}+6\right) \alpha^{3}+ \\
\left(a_{4}-1\right) \alpha^{4}+\left(a_{5}-1\right) \alpha^{5}+a_{6} \alpha^{6}+\cdots+a_{2 \cos } \alpha^{2005} \\
=M\left(\alpha^{2}+3 \alpha-2\right)=0,
\end{array}
$$

where $M$ is the quotient. Therefore,
$$
\begin{array}{l}
a_{0}+a_{1} \alpha+a_{2} \alpha^{2}+\cdots+a_{2} \cos \alpha^{2008} \\
=\alpha^{5}+\alpha^{4}-6 \alpha^{3}+10 \alpha^{2}-4 \alpha \\
=\left(\alpha^{3}-2 \alpha^{2}+2 \alpha\right)\left(\alpha^{2}+3 \alpha-2\right)=0 .
\end{array}
$$