Question

$Provethat\left(\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}\right)={\omega }^2,where\omega isthecuberootofunity$.

Answer 1

$\frac{a+bw+c{w}^2}{c+aw+b{w}^2}$, where $w$ is cube root of unity.

To prove:-

$\frac{a+bw+c{w}^2}{c+aw+b{w}^2}=w^2$

proof

$\frac{a+bw+c{w}^2}{c+aw+b{w}^2}$

$=w^2\frac{(\frac{a}{w^2}+\frac{b}{w}+c)}{(c+aw+b{w}^2)}....\left(i\right)$

now, as $w$ is the cube root of unity

so, $w^3=1$

$\Rightarrow w^3=1$

$\Rightarrow w=\frac{1}{w^2}$ & $w^2=\frac{1}{w}....\left(ii\right)$

reputing the values $\left(ii\right)$ in $\left(i\right)$

$\frac{a+bw+c{w}^2}{c+aw+b{w}^2}=w^2$

$=w^2$

Hence proved