3. Given $\alpha=3+2 \sqrt{2}, \beta=3-2 \sqrt{2}$. If $\alpha^{10}+\beta^{10}$ is a positive integer, then its last digit is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
3. B.

Since $\alpha+\beta=6, \alpha \beta=1$, then $\alpha, \beta$ are the two roots of the equation $x^{2}-6 x+1=0$. Therefore, we have
$$
\left\{\begin{array}{l}
\alpha^{2}=6 \alpha-1, \\
\beta^{2}=6 \beta-1 .
\end{array}\right.
$$

Thus, $\left\{\begin{array}{l}\alpha^{n}=6 \alpha^{n-1}-\alpha^{n-2}, \\ \beta^{n}=6 \beta^{n-1}-\beta^{n-2} .\end{array}\right.$
If we denote $S_{k}=\alpha^{k}+\beta^{k}$, then we have
$$
S_{n}=6 S_{n-1}-S_{n-2} \text {. }
$$

Clearly, $S_{0}=\alpha^{0}+\beta^{0}=2, S_{1}=\alpha+\beta=6$.
Let $\overline{S_{k}}$ represent the last digit of $S_{k}$, then
$$
\overline{S_{n}}=\overline{6} \overline{S_{n-1}}-\overline{S_{n-2}} \text {. }
$$

From equation (2), we get
$$
\begin{array}{l}
\overline{S_{0}}=2, \overline{S_{1}}=6, \overline{S_{2}}=4, \overline{S_{3}}=8, \overline{S_{4}}=4, \overline{S_{5}}=6, \\
\overline{S_{6}}=2, \overline{S_{7}}=6, \overline{S_{8}}=4, \overline{S_{9}}=8, \overline{S_{10}}=4 .
\end{array}
$$