Question

Prove that $\frac{1+2\omega +3{\omega }^2}{2+3\omega +{\omega }^2}+\frac{2+3\omega +{\omega }^2}{3+\omega +2{\omega }^2}=2\omega$.

Answer 1

$1+\omega +{\omega }^2=0$ (sum of cube roots of unity)

$\frac{(1+\omega +{\omega }^2)+(\omega +{2\omega }^2)}{(1+\omega +{\omega }^2)+(1+2\omega )}+\frac{(1+\omega +{\omega }^2)+(1+2\omega )}{(1+\omega +{\omega }^2)+(\omega +{2\omega }^2)}⟹\frac{(\omega +{2\omega }^2)}{(1+2\omega )}+\frac{(1+2\omega )}{(2+{\omega }^2)}⟹\frac{\omega (1+2\omega )}{(1+2\omega )}+\frac{\omega (1+2\omega )}{\omega (2+{\omega }^2)}⟹\omega +\frac{\omega (1+2\omega )}{(2\omega +{\omega }^3)}⟹\omega +\frac{\omega (1+2\omega )}{(1+2\omega )}⟹2\omega$