Question

If $\omega$ is a nonreal cuberoot of unity then $\frac{1+2\omega +3{\omega }^2}{2+3\omega +{\omega }^2}+\frac{2+3\omega +{\omega }^2}{3+\omega +2{\omega }^2}$ is equal to

• A. $1$
• B. $2\omega$
• C. $0$
• D. $-2\omega$

Answer 1

Answer: (B)

$2\omega$

we know $1+\omega +{\omega }^2=0$
$\therefore \frac{1+2\omega +3{\omega }^2}{2+3\omega +{\omega }^2}+\frac{2+3\omega +{\omega }^2}{3+\omega +2{\omega }^2}$
$=\frac{(1+\omega +{\omega }^2)+\omega +2{\omega }^2}{2(1+\omega +{\omega }^2)+(\omega -{\omega }^2)}+\frac{2(1+\omega +{\omega }^2)+(\omega -{\omega }^2)}{3(1+\omega +{\omega }^2)-2\omega -{\omega }^2}$
$=\frac{1+2\omega }{1-\omega }+\frac{1-\omega }{-2-\omega }$
$=\frac{\omega -{\omega }^2}{1-\omega }-\frac{1-\omega }{2+\omega }$
$=\omega -\frac{1-\omega }{1-{\omega }^2}=\omega -\frac{1}{1+\omega }=\omega -\frac{{\omega }^3}{-{\omega }^2}=2\omega ,\because {\omega }^3=1$