Question

If $\omega (\ne 1)$ is cube root of unity satisfying $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^{2}and\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega$ , then the
value of $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ is

• A. 2
• B. -2
• C.
• D. None of these

Answer 1

The correct option is A 2

We have, $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2=\frac{2}{\omega }$
and $\frac{1}{a+{\omega }^2}+\frac{1}{b+\omega 62}+\frac{1}{c+{\omega }^2}=2\omega =\frac{2}{{\omega }^2}$
$\therefore \omega and{\omega }^2$ are roots of the equation
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$
Thus, when x = 1, we get
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{1}=2$