Find $c$ if $a$, $b$, and $c$ are [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$.
Expanding out both sides of the given [equation](https://artofproblemsolving.com/wiki/index.php/Equation) we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$.  Two [complex numbers](https://artofproblemsolving.com/wiki/index.php/Complex_number) are equal if and only if their [real parts](https://artofproblemsolving.com/wiki/index.php/Real_part) and [imaginary parts](https://artofproblemsolving.com/wiki/index.php/Imaginary_part) are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$.  Since $a, b$ are [integers](https://artofproblemsolving.com/wiki/index.php/Integer), this means $b$ is a [divisor](https://artofproblemsolving.com/wiki/index.php/Divisor) of 107, which is a [prime number](https://artofproblemsolving.com/wiki/index.php/Prime_number).  Thus either $b = 1$ or $b = 107$.  If $b = 107$, $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$, but $107^2 + 1$ is not divisible by 3, a contradiction.  Thus we must have $b = 1$, $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive).  Thus $c = 6^3 - 3\cdot 6 = \boxed{198}$.