Find $c$ if $a,b, \; c$ satisfy $c = (a+bi)^3 - 107i$ 
Find $c$ if $a,b, \; c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$

I can try expanding the cube, but that seems too direct. What other ways are there to go about this?

Since $c \in \mathbb Z$, the right-hand side must have an imaginary component of $0$. Expanding, we get
$$c = (a+bi)^3 - 107i = (a^3-3ab^2) + (3a^2b-b^3-107)i$$
Therefore $3a^2b-b^3-107 = 0$, which implies that $(3a^2-b^2)b=107$. Since $107$ is prime, we conclude that $b$ is either $1$ or $107$.
Case 1: $b=1$. Then $3a^2 -1 = 107$, so $a=6$.
Case 2: $b=107$. Then $3a^2 - 107^2 = 1$. But there is no integer solution to this equation.
Therefore the unique solution is $(a,b,c) = (6,1,198)$.
