What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$?
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$
$(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$. Finally, multiply and add to find that the smallest integer higher is $\boxed {\textbf{(C) } 970}$