Question

If $a$ and $b$ are positive integers such that $N=\left(a + i b\right)^{3}-107i$ (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)

$\left(a + i b\right)^{3}-107i=$ Natural number
$\Rightarrow \left(a^{3} - 3 b^{2} a\right)+i\left(3 a^{2} b - b^{3}\right)-107i=$ Natural number
$\Rightarrow 3a^{2}b-b^{3}-107=0$
$\Rightarrow a^{2}=\frac{b^{3} + 107}{3 b}=\frac{b^{2}}{3}+\frac{107}{3 b}$
$\Rightarrow b^{3}+107$ 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 $3$ }