Question

How do you simplify `(p+q*omega+r*omega^2)/(r+p*omega+q*omega^2)` ?

Answer 1

See explanation...

Explanation:

`omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

So `omega^3 = 1` and `omega^4 = omega`

`(p+q*omega+r*omega^2)/(r+p*omega+q*omega^2)`

`=(p*omega^3+q*omega^4+r*omega^2)/(r+p*omega+q*omega^2)`

`=(r*omega^2+p*omega^3+q*omega^4)/(r+p*omega+q*omega^2)`

`=(color(red)(cancel(color(black)((r+p*omega+q*omega^2))))omega^2)/color(red)(cancel(color(black)((r+p*omega+q*omega^2))))`

`=omega^2`