Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$.
1. We start with the given equation:
   \[
   c = (a + bi)^3 - 107i
   \]
   where \(i^2 = -1\).

2. Expand \((a + bi)^3\) using the binomial theorem:
   \[
   (a + bi)^3 = a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3
   \]
   Simplify each term:
   \[
   a^3 + 3a^2(bi) + 3a(b^2i^2) + (b^3i^3)
   \]
   Since \(i^2 = -1\) and \(i^3 = -i\), we get:
   \[
   a^3 + 3a^2(bi) + 3a(b^2(-1)) + b^3(-i)
   \]
   Simplify further:
   \[
   a^3 + 3a^2bi - 3ab^2 - b^3i
   \]
   Combine the real and imaginary parts:
   \[
   (a^3 - 3ab^2) + (3a^2b - b^3)i
   \]

3. Substitute this back into the original equation:
   \[
   c = (a^3 - 3ab^2) + (3a^2b - b^3)i - 107i
   \]
   Combine the imaginary parts:
   \[
   c = (a^3 - 3ab^2) + (3a^2b - b^3 - 107)i
   \]

4. For \(c\) to be a positive integer, the imaginary part must be zero:
   \[
   3a^2b - b^3 - 107 = 0
   \]
   Rearrange to solve for \(b\):
   \[
   b(3a^2 - b^2) = 107
   \]

5. Since \(107\) is a prime number, \(b\) must be either \(1\) or \(107\):
   - If \(b = 1\):
     \[
     1(3a^2 - 1^2) = 107
     \]
     Simplify:
     \[
     3a^2 - 1 = 107
     \]
     \[
     3a^2 = 108
     \]
     \[
     a^2 = 36
     \]
     \[
     a = 6
     \]
     So, \(a = 6\) and \(b = 1\).

   - If \(b = 107\):
     \[
     107(3a^2 - 107^2) = 107
     \]
     Simplify:
     \[
     3a^2 - 107^2 = 1
     \]
     \[
     3a^2 - 11449 = 1
     \]
     \[
     3a^2 = 11450
     \]
     \[
     a^2 = \frac{11450}{3}
     \]
     Since \(\frac{11450}{3}\) is not an integer, there is no solution for \(b = 107\).

6. Therefore, the only solution is \(a = 6\) and \(b = 1\). Substitute these values back into the expression for \(c\):
   \[
   c = a(a^2 - 3b^2)
   \]
   \[
   c = 6(6^2 - 3 \cdot 1^2)
   \]
   \[
   c = 6(36 - 3)
   \]
   \[
   c = 6 \cdot 33
   \]
   \[
   c = 198
   \]

The final answer is \(\boxed{198}\).