Question
If $\omega$ is a complex cube root of unity, show that : $\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^2\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^2}=\omega^2$

Answer

\begin{aligned}
& \text { L.H.S. }=\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^2\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^2} \\
& =\frac{a \omega^3+b \omega^4+c \omega^2}{c+a \omega+b \omega^2} \ldots \ldots . .\left[\omega^3=1, \omega^4=\omega\right] \\
& =\frac{\omega^2\left(c+a \omega+b \omega^2\right)}{c+a \omega+b \omega^2} \\
& =\omega^2 \\
& =\text { R.H.S. }
\end{aligned}

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