Question

If $\omega$ is a non-real cube root of unity, then
$\frac{1+2\omega+3\omega^{2}}{2+3\omega+\omega^{2}} + \frac{2+3\omega +3\omega ^{2}}{3+3\omega +2\omega ^{2}}$ is equal to

• A. $-2\,\omega$
• B. $2\,\omega$
• C. $\omega$
• D. $0$

Answer 1

Answer: (B)

We have, $\frac{1+2\omega+3\omega^{2}}{2+3\omega+\omega^{2}} + \frac{2+3\omega +3\omega ^{2}}{3+3\omega +2\omega ^{2}}$
We know that $\omega^{3} = 1$
$\frac{\omega^{3}+2\omega +3\omega ^{2}}{2+3\omega +\omega ^{2}} + \frac{2\omega^{3}+3\omega +3\omega ^{2}}{3+3\omega +2\omega ^{2}}$
$= \frac{\omega\left(\omega^{2}+2 +3\omega\right)}{2+3\omega +\omega ^{2}} + \frac{\omega\left(2\omega^{2}+3 +3\omega\right)}{\left(3+3\omega +2\omega ^{2}\right)} = \omega+\omega = 2\,\omega$