Question

If $\omega$ and ${\omega }^2$ are the nonreal cube roots of unity and $\left[1/\right(a+\omega \left)\right]+\left[1/\right(b+\omega \left)\right]+\left[1/\right(c+\omega \left)\right]=2{\omega }^2$ and $\left[1/\right(a+\omega {)}^2]+[1/(b+\omega {)}^2]+\left[1/\right(c+\omega {)}^2]=2\omega$, then find the value of $\left[1/\right(a+1\left)\right]+\left[1/\right(b+1\left)\right]+\left[1/\right(c+1\left)\right]$.

• A. $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=0$
• B. $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$
• C. $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$
• D. $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=-2$

Answer 1

The correct option is C $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$

The given relations can be rewritten as $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=\frac{2}{\omega }$
and $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=\frac{2}{{\omega }^2}$
$⟹\omega$ and ${\omega }^2$ are roots of $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$
$⟹\frac{3x^2+2(a+b+c)x+bc+ca+ab}{(a+x)(b+x)(c+x)}=\frac{2}{x}$
$⟹x^3+(bc+ca+ab)x-2abc=0$ (1)
Two roots of (1) are $\omega$ and ${\omega }^2$. Let the third root be $\alpha$. Then, $\alpha +\omega +{\omega }^2=0$ or $\alpha =-\omega -{\omega }^2=1$
Therefore, $\alpha =1$ will satisfy Eq. (1). Hence,
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$
Ans: C