Question

Show that: $\frac{(a+bw+c{w}^2)}{c+aw+b{w}^2}=w^2$

Answer 1

$\left(\frac{a+b\omega +c{\omega }^2}{c+a\omega +b\omega }\right)\times \left(\frac{\omega }{\omega }\right)=\left(\frac{a\omega +b{\omega }^2+c{\omega }^3}{c+a\omega +b{\omega }^2}\right)\times \left(\frac{1}{\omega }\right)=\left(\frac{c+a\omega +b{\omega }^2}{c+a\omega +b{\omega }^2}\right)\times \left(\frac{1}{\omega }\right)=\left(\frac{1}{\omega }\right)\times \left(\frac{\omega }{{\omega }^2}\right)={\omega }^2$