Question

Prove that $\frac{1}{1+2\omega }+\frac{1}{2+\omega }-\frac{1}{1+\omega }=0$ Where $\omega$is imaginary cube root of unity.

Answer 1

$1+\omega +{\omega }^2=0$ $\therefore 1+\omega =-{\omega }^2$
$\therefore 1+2\omega =\omega -{\omega }^2$
$2+\omega =1-{\omega }^2,1+\omega =-{\omega }^2$
$\therefore L.H.S=\frac{1}{\omega (1-\omega )}+\frac{1}{(1-\omega )(1+\omega )}+\frac{1}{{\omega }^2}$
$=\frac{\omega (1+\omega )+{\omega }^2+1-{\omega }^2}{{\omega }^2(1-{\omega }^2)}=\frac{1+\omega +{\omega }^2}{{\omega }^2(1-{\omega }^2)}=0$