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 ${\omega }^3=1$
$\frac{{\omega }^3+2\omega +3{\omega }^2}{2+3\omega +{\omega }^2}+\frac{2{\omega }^3+3\omega +3{\omega }^2}{3+3\omega +2{\omega }^2}$
$=\frac{\omega \left({\omega }^2+2+3\omega \right)}{2+3\omega +{\omega }^2}+\frac{\omega \left(2{\omega }^2+3+3\omega \right)}{\left(3+3\omega +2{\omega }^2\right)}=\omega +\omega =2\omega$