4-232 If $a, b, c$ are positive integers, satisfying $c=(a+b i)^{3}-107 i$, find $c$ (where $i^{2}=-1$).


[Solution] From the given information, we have
$$c=\left(a^{3}-3 a b^{2}\right)+i\left(3 a^{2} b-b^{3}-107\right) .$$

Since $c$ is a positive integer, it follows that
$$3 a^{2} b-b^{3}-107=0$$

which simplifies to $b\left(3 a^{2}-b^{2}\right)=107$.
Since $a, b$ are positive integers, and 107 is a prime number, there can only be the following two scenarios.
Scenario $1\left\{\begin{array}{l}b=107, \\ 3 a^{2}-b^{2}=1 .\end{array}\right.$ In this case, $\quad 3 a^{2}=107^{2}+1$.
However, the left side of the equation is a multiple of 3, while the right side is not a multiple of 3, so the equation cannot hold. In other words, Scenario 1 cannot occur.

Scenario $2\left\{\begin{array}{l}b=1, \\ 3 a^{2}-b^{2}=107 .\end{array}\right.$ In this case, $3 a^{2}=108$, solving for $a$ gives $a=6$.
Then, $\quad c=a^{3}-3 a b^{2}=6^{3}-3 \times 6 \times 1^{2}=198$