Question

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

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

Answer 1

Answer: (B)

$2\omega$

We know,

${\omega }^3=1$ and

$1+\omega +{\omega }^2=0$

Now,

$\frac{1+2\omega +3{\omega }^2}{2+3\omega +{\omega }^2}+\frac{2+3\omega +{\omega }^2}{3+\omega +2{\omega }^2}$

$\Rightarrow$ $\frac{(1+\omega +{\omega }^2)+(\omega +{\omega }^2)+{\omega }^2}{(1+\omega +{\omega }^2)+(1+\omega )+\omega }+\frac{(1+\omega +{\omega }^2)+(1+\omega )+\omega }{(1+\omega +{\omega }^2)+(1+{\omega }^2)+1}$

$\Rightarrow$ $\frac{0+(-1)+{\omega }^2}{-{\omega }^2+\omega }+\frac{0+(-{\omega }^2)+\omega }{0+(-\omega )+1}$

$\Rightarrow$ $\frac{(\omega -1)(\omega +1)}{-\omega (\omega -1)}+\frac{\omega (1-\omega )}{1-\omega }$

$\Rightarrow$ $\frac{-(\omega +1)}{\omega }+\omega$

$\Rightarrow$ $\frac{-(\omega +1)+{\omega }^2}{\omega }$

$\Rightarrow$ $\frac{-(-{\omega }^2)+{\omega }^2}{\omega }$

$\Rightarrow$ $\frac{2{\omega }^2}{\omega }$

$\Rightarrow$ $2\omega$