Welcome 2016! Part 22

Expanding both sides of given equation, we have

$c+107i=(a^3-3ab^2)+(3a^2b-b^3 )i$

Two complex numbers are equal if and only if their real parts and imaginary parts are equal. So,

$c=a^3-3ab^2 \ \text{and} \ 107=3a^2b-b^3=(3a^2-b^2)b$

Since $\text{a and b}$ are integers, this means that b is a divisor of 107, which is a prime number. Thus either $b=1$ or $b=107$ .

If $b=107$ , $3a^2-107^2=1$ . So, $3a^2=107^2+1$ . But $107^2+1$ is not divisible by 3.

Thus, we must have $b=1$ , $3a^2=108$ . So, $a^2=36$ , $a=6$ (Since we know $a$ is positive ).

At last,

$c=6^3-3 \times 6 \ , \ c=198$