$13 \cdot 14$ is greater than $(\sqrt{3}+\sqrt{2})^{6}$ by the smallest integer
(A) 972.
(B) 971.
(C) 970.
(D) 969.
(E) 968.
(26th American High School Mathematics Examination, 1975)
[Solution] Since $a^{6}+b^{6}=\left(a^{2}+b^{2}\right)\left(a^{4}-a^{2} b^{2}+b^{4}\right)$
$$
=\left(a^{2}+b^{2}\right)\left[\left(a^{2}+b^{2}\right)^{2}-3 a^{2} b^{2}\right],
$$

Notice that
$$
\begin{aligned}
& (\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6} \\
= & {\left[(\sqrt{3}+\sqrt{2})^{2}+(\sqrt{3}-\sqrt{2})^{2}\right]\left\{\left[(\sqrt{3}+\sqrt{2})^{2}+(\sqrt{3}-\sqrt{2})^{2}\right]^{2}\right.} \\
& \left.-3(\sqrt{3}+\sqrt{2})^{2}(\sqrt{3}-\sqrt{2})^{2}\right\} \\
= & 10 \cdot\left(10^{2}-3\right)=970 .
\end{aligned}
$$

Since $\sqrt{3}-\sqrt{2}<1$, it follows that $(\sqrt{3}-\sqrt{2})^{6}<1$.
Thus, the smallest integer greater than $(\sqrt{3}+\sqrt{2})^{6}$ is 970.
Therefore, the answer is $(C)$.