Question

$1,\omega ,{\omega }^2$ are cube roots of unity then
$\frac{1}{1+2\omega }-\frac{1}{1+\omega }+\frac{1}{2+\omega }=$

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

Answer 1

The correct option is A $0$

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

$\left[\begin{array}{cc}\because \omega =1& \\ \Rightarrow {\omega }^2=\frac{1}{\omega }& \end{array}\right]$

$-\frac{1}{1+\omega }=\frac{1}{{\omega }^2}=\omega [\because {\omega }^3=1]$

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

Adding $\left(1\right)+\left(2\right)+\left(3\right)$

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

$=0$