75 is greater than $(\sqrt{3}+\sqrt{2})^{6}$, the smallest integer is
A. 972
B. 971
C. 970
D. 969
75 C. Since $0<\sqrt{3}-\sqrt{2}<1$,
$$
(\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}=970 .
$$

Therefore, 970 is the smallest integer greater than $(\sqrt{3}+\sqrt{2})^{6}$.