If $$1,\omega ,{\omega ^2}$$  are the three cube roots of unity, then what is $$\frac{{\left( {a{\omega ^6} + b{\omega ^4} + c{\omega ^2}} \right)}}{{\left( {b + c{\omega ^{10}} + a{\omega ^8}} \right)}}$$    equal to ?

Solution :
$$1,\omega $$  and $$\omega^2 $$ are the three cube roots of unity.
$$ \Rightarrow 1 + \omega + {\omega ^2} = 0{\text{ and }}{\omega ^3} = 1.$$
The given expression
$$\eqalign{ & \frac{{a{\omega ^6} + b{\omega ^4} + c{\omega ^2}}}{{b + c{\omega ^{10}} + a{\omega ^8}}} = \frac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}\left[ {{\omega ^6} = 1,{\omega ^4} = \omega } \right] \cr & = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\omega \left( {b + c\omega + a{\omega ^2}} \right)}}\left[ {{\text{Multiplying }}{N^r}{\text{ and }}{D^r}{\text{ by }}\omega {\text{.}}} \right] \cr & = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {a{\omega ^3} + b\omega + c{\omega ^2}} \right)}} = \frac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {a + b\omega + c{\omega ^2}} \right)}} = \omega \cr} $$