Question

If for three non-zero and unequal real numbers $a,b$ and $c$; $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2$ and $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega$, where ${\omega }^2$ and $\omega$ are the complex cube roots of unity, then $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=$

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is C

2

The given equations can be written as

$\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2=\frac{2}{\omega }$ ... (1) [${\omega }^3=1\Rightarrow {\omega }^2=\frac{1}{\omega }$]

$\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega =\frac{2}{{\omega }^2}$ ... (2) [${\omega }^3=1\Rightarrow \omega =\frac{1}{{\omega }^2}$]

Now consider the cubic equation

$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$

From (1) and (2) we can deduce that $\omega$ and ${\omega }^2$ are solution of this cubic equation with real coefficient using $a,b$ and $c.$ Hence, $x=1$ must be its third root.

$\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2}{1}=2$