Question

If $\omega$ is complex cube root of unity and a, b, c are three real numbers such that
$\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2$ 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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Since, $w^2=\frac{1}{w}$ and $w=\frac{1}{w^2}$
Given relation may be rewritten as
$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}=\frac{2}{w}$ and $\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}=\frac{2}{w^2}$
Clearly $w$ and $w^2$ are the roots of
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$
$\Rightarrow \frac{(b+x)(c+x)+(a+x)(c+x)+(a+x)(b+x)}{(a+x)(b+x)(c+x)}=\frac{2}{x}$
$\Rightarrow x[3x^2+2(a+b+c)x+bc+ca+ab]=2[abc+(bc+ca+ab)x+(a+b+c)x^2+x^3]$
$\Rightarrow x^3-(bc+ca+ab)x-2abc=0$
If $\alpha$ is the third root of this equation then sum of the roots
$\alpha +w+w^2=0\Rightarrow \alpha =1$
Hence, $1$ is the root of equation, we get
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$