solver integers a, b, and c in (a+b*i)^3 -107i = c

solver integers a, b, and c in (a+b*i)^3 -107i = c
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1 Answer

(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.

by Top Rated User (1.2m points)

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