Question

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

Answer 1

$\frac{1}{1+2\omega }+\frac{1}{2+\omega }-\frac{1}{1+\omega }$

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

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

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

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

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

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

$=\frac{1}{1+\omega }-\frac{1}{1+\omega }$

$=0$.