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 $a$ exist, but only finitely many;
(C) such $a$ do not exist,
(D) none of the conclusions (A),
(B),
(C) are correct.
For any given natural number $n$, there must be $n^{6}+3 \cdot n^{2} \cdot 3 \cdot\left(n^{2}+3\right)+3^{3}=\left(n^{2}+3\right)^{3}$. Let $a=3 n^{2}\left(n^{2}+3\right)+3^{2}$, which makes $n^{6}+3 a$ a cube of a positive integer, so such an $a$ exists.

And it is easy to find that $\left(n^{2}+3 k\right)^{3}=n^{6}+3 \cdot n^{2} \cdot 3 k$ $\cdot\left(n^{2}+3 k\right)+(3 k)^{3}$. We only need to let $a_{k}=3 n^{2} k\left(n^{2}+\right.$ $3 k)+9 k^{3}(k=1,2,3, \cdots)$, all of which satisfy $n^{8}+$ $3 a_{k}$ being a cube of a positive integer (where $a_{k}$ are all positive integers), meaning such an $a$ exists and there are infinitely many. Therefore, the answer should be $(A)$.