1. Given $a=\frac{-\sqrt{6}+\sqrt{6-4 \sqrt{2}}}{2}$. Then $a^{3}+$ $\sqrt{6} a^{2}+\sqrt{2} a+\sqrt{6}$ is ( ).
(A) $-\sqrt{2}$
(B) $\sqrt{2}$
(C) $-\sqrt{6}$
(D) $\sqrt{6}$
-、1.D.
Notice that $a=\frac{-\sqrt{6}+\sqrt{\sqrt{6}^{2}-4 \times 1 \times \sqrt{2}}}{2 \times 1}$, then $a$ is a root of the equation $x^{2}+\sqrt{6} x+\sqrt{2}=0$.
Therefore, $a^{2}+\sqrt{6} a+\sqrt{2}=0$.
Thus, $a^{3}+\sqrt{6} a^{2}+\sqrt{2} a+\sqrt{6}$
$$
=a\left(a^{2}+\sqrt{6} a+\sqrt{2}\right)+\sqrt{6}=\sqrt{6} .
$$