Q.If $\omega$ is a complex cube root of unity, show that $\dfrac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2} = \omega^2$

The cube roots of unity are the three solutions of $x^3=1$, found by factoring $x^3-1=(x-1)(x^2+x+1)=0$: one real root $x=1$, and two complex conjugate roots coming from the quadratic factor, $\omega=-\dfrac12+\dfrac{\sqrt3}{2}i$ and $\omega^2=-\dfrac12-\dfrac{\sqrt3}{2}i$ (so $\omega$ and $\omega^2=\bar\omega$ are complex conjugates of each other). These three roots — $1,\omega,\omega^2$ — satisfy a standing toolkit of identities used constantly in this chapter's exercises: $\omega^3=1$ (so powers of $\omega$ cycle every 3 steps, exactly like powers of $i$ cycle every 4), the sum identity $1+\omega+\omega^2=0$ (equivalently $\omega^2+\omega+1=0$, so $\omega^2=-1-\omega$ and $\omega+1=-\omega^2$), the reciprocal identities $\dfrac1\omega=\omega^2$ and $\dfrac{1}{\omega^2}=\omega$, and the periodicity rules $\omega^{3n}=1$, $\omega^{3n+1}=\omega$, $\omega^{3n+2}=\omega^2$ for any integer $n$. Also, $1=e^{2\pi i}$, $\omega=e^{2\pi i/3}$, and $\omega^2=e^{4\pi i/3}$, linking the cube roots of unity to the exponential form and De Moivre's theorem. These identities let complicated-looking expressions in $\omega$ (like $(1+\omega-\omega^2)^6$ or $(a+b\omega+c\omega^2)$) collapse to small integers or simple multiples of $\omega,\omega^2$, and the same toolkit works for a …