Question

If $\omega$, ${\omega }^2$ are imaginary cube roots of unity and

$\frac{1}{a+\omega }$ + $\frac{1}{b+\omega }$ + $\frac{1}{c+\omega }$ = 2${\omega }^2$ and $\frac{1}{a+{\omega }^2}$ + $\frac{1}{b+{\omega }^2}$ + $\frac{1}{c+{\omega }^2}$ = 2$\omega$, then $\frac{1}{a+1}$ + $\frac{1}{b+1}$ + $\frac{1}{c+1}$ is equal to:

• A. 1
• B. -1
• C. 3
• D. 2

Answer 1

The correct option is D

2

Form the given condition, $\omega$ and ${\omega }^2$ are roots of the equation

$\frac{1}{a+x}$ + $\frac{1}{b+x}$ + $\frac{1}{c+x}$ = $\frac{2}{x}$[ ∵ ${\omega }^2=\frac{1}{\omega }$, $\omega$ = $\frac{1}{{\omega }^2}$]

$\Rightarrow$ $x^3$ - (ab + bc + ca)x - 2abc = 0.

Let $\alpha$ be the third root. Then $\alpha$ + $\omega$ + ${\omega }^2$ = 0

i.e., $\alpha$ = 1. Hence $\frac{1}{a+1}$ + $\frac{1}{b+1}$ + $\frac{1}{c+1}$ = 2.