How many positive integers $n$ are there such that $3n^2 + 3n + 7$ is a perfect cube?

$ 
\textbf{a)}\ 0
\qquad\textbf{b)}\ 1
\qquad\textbf{c)}\ 3
\qquad\textbf{d)}\ 7
\qquad\textbf{e)}\ \text{Infinitely many}
$
To determine how many positive integers \( n \) exist such that \( 3n^2 + 3n + 7 \) is a perfect cube, we start by setting up the equation:

\[ 3n^2 + 3n + 7 = y^3 \]

We need to analyze this equation modulo 3. First, we consider the expression \( 3n^2 + 3n + 7 \) modulo 3:

\[ 3n^2 + 3n + 7 \equiv 0n^2 + 0n + 7 \equiv 7 \equiv 1 \pmod{3} \]

This implies that \( y^3 \equiv 1 \pmod{3} \). The cubic residues modulo 3 are 0, 1, and 2. We need to check which values of \( y \) satisfy \( y^3 \equiv 1 \pmod{3} \):

- If \( y \equiv 0 \pmod{3} \), then \( y^3 \equiv 0 \pmod{3} \)
- If \( y \equiv 1 \pmod{3} \), then \( y^3 \equiv 1 \pmod{3} \)
- If \( y \equiv 2 \pmod{3} \), then \( y^3 \equiv 8 \equiv 2 \pmod{3} \)

Thus, \( y \equiv 1 \pmod{3} \) is the only possibility. Let \( y = 3k + 1 \) for some integer \( k \). Substituting \( y = 3k + 1 \) into the original equation, we get:

\[ 3n^2 + 3n + 7 = (3k + 1)^3 \]

Expanding the right-hand side:

\[ (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 \]

Thus, we have:

\[ 3n^2 + 3n + 7 = 27k^3 + 27k^2 + 9k + 1 \]

Rearranging terms, we get:

\[ 3n^2 + 3n + 7 - 1 = 27k^3 + 27k^2 + 9k \]

\[ 3n^2 + 3n + 6 = 27k^3 + 27k^2 + 9k \]

Dividing both sides by 3:

\[ n^2 + n + 2 = 9k^3 + 9k^2 + 3k \]

Rewriting:

\[ n^2 + n + 2 = 3(3k^3 + 3k^2 + k) \]

This implies:

\[ n^2 + n + 2 \equiv 0 \pmod{3} \]

We now check the possible values of \( n \) modulo 3:

- If \( n \equiv 0 \pmod{3} \), then \( n^2 + n + 2 \equiv 0 + 0 + 2 \equiv 2 \pmod{3} \)
- If \( n \equiv 1 \pmod{3} \), then \( n^2 + n + 2 \equiv 1 + 1 + 2 \equiv 4 \equiv 1 \pmod{3} \)
- If \( n \equiv 2 \pmod{3} \), then \( n^2 + n + 2 \equiv 4 + 2 + 2 \equiv 8 \equiv 2 \pmod{3} \)

In all cases, \( n^2 + n + 2 \not\equiv 0 \pmod{3} \). Therefore, there are no positive integers \( n \) such that \( 3n^2 + 3n + 7 \) is a perfect cube.

The final answer is \(\boxed{0}\)