What is the maximum number of primes that divide both the numbers $n^3+2$ and $(n+1)^3+2$ where $n$ is a positive integer?

$ 
\textbf{(A)}\ 3
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 1
\qquad\textbf{(D)}\ 0
\qquad\textbf{(E)}\ \text{None of above}
$
1. Let \( a \) be a prime that divides both \( n^3 + 2 \) and \( (n+1)^3 + 2 \). Then, \( a \) must divide the difference of these two expressions:
   \[
   a \mid ((n+1)^3 + 2) - (n^3 + 2)
   \]
   Simplifying the difference, we get:
   \[
   (n+1)^3 + 2 - (n^3 + 2) = (n^3 + 3n^2 + 3n + 1 + 2) - (n^3 + 2) = 3n^2 + 3n + 1
   \]
   Therefore, \( a \mid 3n^2 + 3n + 1 \).

2. Additionally, \( a \) must divide \( (n+1)^3 + 2 \) and \( n^3 + 2 \). Consider the expression:
   \[
   (n+1)^3 + 2 - (3n^2 + 3n + 1) = n^3 + 3n^2 + 3n + 1 + 2 - 3n^2 - 3n - 1 = n^3 + 2
   \]
   This implies:
   \[
   a \mid n^3 + 2
   \]
   and
   \[
   a \mid (n+1)^3 + 2
   \]

3. Now, we need to find the greatest common divisor (gcd) of \( n^3 + 2 \) and \( (n+1)^3 + 2 \). We already know:
   \[
   \gcd(n^3 + 2, (n+1)^3 + 2) = \gcd(n^3 + 2, 3n^2 + 3n + 1)
   \]

4. To simplify further, consider the gcd of \( n^3 + 2 \) and \( 4 \):
   \[
   \gcd(n^3 + 2, 4)
   \]
   We need to evaluate \( n^3 \mod 4 \). The possible values of \( n \mod 4 \) are 0, 1, 2, and 3:
   - If \( n \equiv 0 \pmod{4} \), then \( n^3 \equiv 0 \pmod{4} \) and \( n^3 + 2 \equiv 2 \pmod{4} \).
   - If \( n \equiv 1 \pmod{4} \), then \( n^3 \equiv 1 \pmod{4} \) and \( n^3 + 2 \equiv 3 \pmod{4} \).
   - If \( n \equiv 2 \pmod{4} \), then \( n^3 \equiv 8 \equiv 0 \pmod{4} \) and \( n^3 + 2 \equiv 2 \pmod{4} \).
   - If \( n \equiv 3 \pmod{4} \), then \( n^3 \equiv 27 \equiv 3 \pmod{4} \) and \( n^3 + 2 \equiv 1 \pmod{4} \).

   Therefore, the possible values of \( n^3 + 2 \mod 4 \) are 2, 3, and 1. The gcd of any of these values with 4 is either 1 or 2.

5. Since \( \gcd(n^3 + 2, 4) \) can be 1 or 2, we conclude that \( a \) can be at most 2. However, since \( a \) must be a prime number, the only possible prime is 2.

6. Therefore, the maximum number of primes that can divide both \( n^3 + 2 \) and \( (n+1)^3 + 2 \) is 1.

The final answer is \(\boxed{1 = C}\)