Question

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

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is B $2$

Since, $w^2=\frac{1}{w}$ and $w=\frac{1}{w^2}$, we get $w$ and $w^2=$ are roots of $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$ ...(1)
$\Rightarrow \frac{(b+x)(c+x)+(a+x)(c+x)+(a+x)(b+x)}{(a+x)(b+x)(c+x)}=\frac{2}{x}$
$\Rightarrow x[3x^2+2(a+b+c)x+(bc+ca+ab\left)\right]$
$=2[abc+(bc+ca+ab)x+(a+b+c)x^2+x^3]$
$\Rightarrow x^3-(bc+ca+ab)x-2abc=0$
Note $w$,$w^2$ are the roots of this equation.
If $\alpha$ is the third root of this equation, then,
$\alpha +w+w^2=0$ or $\alpha =1$
Since, $1$ is a root of (1) we get,
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$