How many triples of positive integers $(a,b,c)$ are there such that $a!+b^3 = 18+c^3$?

$ 
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 1
\qquad\textbf{(E)}\ 0
$
1. We start by analyzing the given equation \(a! + b^3 = 18 + c^3\). We need to find the number of triples \((a, b, c)\) of positive integers that satisfy this equation.

2. First, consider the equation modulo 7. We know that the possible values of \(c^3 \mod 7\) are \(-1, 0, 1\) because the cubes of integers modulo 7 can only be these values. Therefore, \(18 + c^3 - b^3 \mod 7\) can be \(2, 3, 4, 5, 6 \mod 7\).

3. Next, consider \(a!\) modulo 7. For \(a \geq 7\), \(a!\) is divisible by 7, so \(a! \equiv 0 \mod 7\). Therefore, \(a\) must be less than 7 for \(a! + b^3 = 18 + c^3\) to hold.

4. We now check the values of \(a\) from 1 to 6:
   - For \(a = 1\), \(1! = 1\). The equation becomes \(1 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = 17\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = 17\).
   - For \(a = 2\), \(2! = 2\). The equation becomes \(2 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = 16\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = 16\).
   - For \(a = 3\), \(3! = 6\). The equation becomes \(6 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = 12\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = 12\).
   - For \(a = 4\), \(4! = 24\). The equation becomes \(24 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = -6\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = -6\).
   - For \(a = 5\), \(5! = 120\). The equation becomes \(120 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = -102\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = -102\).
   - For \(a = 6\), \(6! = 720\). The equation becomes \(720 + b^3 = 18 + c^3\). Rearranging, we get \(b^3 - c^3 = -702\). There are no integer solutions for \(b\) and \(c\) such that \(b^3 - c^3 = -702\).

5. We have checked all possible values of \(a\) from 1 to 6 and found that none of them satisfy the equation \(a! + b^3 = 18 + c^3\).

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