Question

If $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }+\frac{1}{d+\omega }=\frac{1}{\omega }$, where $a,b,c,d\in$ $R$ and $\omega$ is a cube root of unity then $\Sigma \frac{3}{a^2-a+1}$ is equal to

• A. 1
• B. 2
• C. 3
• D. $\frac{1+}{3}$

Answer 1

Answer: (C)

Given that
$\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }+\frac{1}{d+\omega }=\frac{1}{\omega }\dots (i)$
Taking conjugate we get,
$\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}+\frac{1}{d+{\omega }^2}=\frac{1}{{\omega }^2}\dots (ii)$
Subtract equation (ii) from equation (i) we get
$\Sigma \left(\frac{1}{a+\omega }\frac{1}{a+{\omega }^2}\right)=\frac{1}{\omega }-\frac{1}{{\omega }^2}$
$\Rightarrow \Sigma \frac{1}{(a+\omega )\left(a+{\omega }^2\right)}={\omega }^2-\omega$
So $\Sigma \frac{1}{a^2-a+1}=1\Sigma \frac{3}{a^2-a+1}=3$