(a+ib)³=a³+3a²ib-3ab²-ib³=
a³-3ab²+i(3a²b-b³). This has to be equal to:
c+107i, so we equate real and imaginary parts:
c=a³-3ab² and 3a²b-b³=107.
c=a(a²-3b²) and b(3a²-b²)=107.
107 is a prime number so b=1 and 3a²=107+1=108, a²=36, a=±6.
Therefore, c=±6(36-3)=±198.
(i±6)³-107i=±198.