Question

If $a$ and $b$ are positive integers such that $N={\left(a+ib\right)}^3-107i$ (where $N$ is a natural number), then the value of $a$ is equal to (where $i^2=-1$ )

• A. $4$
• B. $5$
• C. $6$
• D. $9$

Answer 1

Answer: (C)

${\left(a+ib\right)}^3-107i=$ Natural number
$\Rightarrow \left(a^3-3b^{2}a\right)+i\left(3a^{2}b-b^3\right)-107i=$ Natural number
$\Rightarrow 3a^{2}b-b^3-107=0$
$\Rightarrow a^2=\frac{b^3+107}{3b}=\frac{b^2}{3}+\frac{107}{3b}$
$\Rightarrow b^3+107$ is a multiple of $3$ and $b$ is a factor of $107$
$\Rightarrow b=1$ and $a^2=\frac{108}{3}=36\Rightarrow a=6$
{ $b$ cannot be equal to $107$ , because ${\left(107\right)}^3+107$ is not a multiple of $3$ }