1. For any given natural number $n$, if $n^{6}+3 a$ is a cube of a positive integer, where $a$ is a positive integer. Then
A. There are infinitely many such $a$
B. Such $a$ exists, but there are only finitely many
C. Such $a$ does not exist
D. None of the conclusions A, B, C are correct
Solve:
$$
\left(n^{2}+3 k\right)^{3}=n^{6}+27 k^{3}+9 n^{2} k\left(n^{2}+3 k\right)=n^{6}+3\left[9 k^{3}+3 n^{2} k\left(n^{2}+3 k\right)\right] \text {, }
$$

Take $a=9 k^{3}+3 n^{2} k\left(n^{2}+3 k\right)\left(k \in \mathbf{N}^{*}\right)$ which satisfies the problem. Therefore, the answer is $A$.