Question

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

Answer 1

Consider $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$ ...(1)
$\omega$ & ${\omega }^2$ are the roots of the eq. (1)
Rewriting eq. (1) as $x^3-x(ab+bc+ac)-2abc=0$
Let $y$ be its third root.
Now, sum of roots $=\omega +{\omega }^2+y=0$
Therefore, $y=1$
Substituting in (1), we get $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$
Ans: 2