Question

If $a$ and $b$ are positive integers such that $N=(a+ib{)}^3-107i$ is a positive integer then $\frac{N}{6}$ is

Answer 1

$N=(a+ib{)}^3-107i$
$=a^3-i{b}^3+3a^{2}ib-3a{b}^2-107i$
$=(-b^3+3a^{2}b-107)i+a^3-3a{b}^2$
Since, $N$ is a positive integer so,
$Im\left(z\right)=0$
$\Rightarrow -b^3+3a^{2}b-107=0\Rightarrow b(3a^2-b^2)=107$
$\therefore b=1\text{and}3a^2-b^2=107\Rightarrow a=6$
$N=a^3-3a{b}^2=198$
$\therefore \frac{N}{6}=33$