Question

$\frac{(a+b\omega +c{\omega }^2)}{c+a\omega +b{\omega }^2}={\omega }^2$

Answer 1

As $1,\omega ,{\omega }^2$ are roots of $1$
$1+\omega +{\omega }^2=0\to 1$
${\omega }^3=1\to 2$
Now, $a+b\omega +c{\omega }^2=a{\omega }^3+b\omega +c{\omega }^2$
$=b\omega +c{\omega }^2+a{\omega }^3$
$=\omega (b+c\omega +a{\omega }^2)$
$\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}=\omega$
And $a+b\omega +c{\omega }^2=a+b\omega +\frac{c}{\omega }$ (Here ${\omega }^2=\frac{1}{\omega }$)
$\frac{(a+b\omega +c{\omega }^2)}{c+a\omega +b{\omega }^2}=\frac{1}{\omega }={\omega }^2$