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 $\Sigma \frac{1}{a^{2}-a+1}=1 \Sigma \frac{3}{a^{2}-a+1}=3$