1. The integer part of the real number $\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}$ is ( ).
(A) 969
(B) 970
(C) 971
(D) None of the above conclusions is correct
$-1 . \mathrm{A}$.
$$
\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}=(\sqrt{3}+\sqrt{2})^{6} \text {. }
$$

Let $\sqrt{3}+\sqrt{2}=a, \sqrt{3}-\sqrt{2}=b$. Then
$$
a+b=2 \sqrt{3}, a b=1 \text {. }
$$

Thus, $a^{2}+b^{2}=(a+b)^{2}-2 a b=10$,
$$
a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)=18 \sqrt{3} \text {. }
$$

Therefore, $(\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}$
$$
\begin{array}{l}
=a^{6}+b^{6}=\left(a^{3}+b^{3}\right)^{2}-2 a^{3} b^{3} \\
=(18 \sqrt{3})^{2}-2 \times 1^{3}=970 .
\end{array}
$$

Since $0<\sqrt{3}-\sqrt{2}<1$, then $0<(\sqrt{3}-\sqrt{2})^{6}<1$.
Thus, $969<(\sqrt{3}+\sqrt{2})^{6}<970$.
Therefore, the integer part of $\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}$ is
969.