Problem: Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$. Solution: This is one of those problems where you sit for a minute and think "What!? How on earth do we find $c$ with such limited information?" Then, you realize components must match up with complex numbers. Namely, $ c+107i=(a+bi)^3 = (a^3-3ab^2)+(3a^2b-b^2)i \implies 107 = b(3a^2-b^2) .$ From here, you might see that $107$ is prime, so either $b = 1$ or $b = 107.$ If $b = 107,$ then $107^2 = 3a^2$ which is not possible since $ 3 \not\mid 107^2.$ Thus, $b = 1$ and we get that $a = 6$. Thus, $c = (a^3-3ab^2) = 198.$