Question

If ω is a cube root of unity show that (1-ω+ω²)(1+ω-ω²)=4

Answer 1

As $\omega$ is a cube root of unity, $\omega^3 = 1$. Hence

$$0 = 1^3 - \omega^3 = (1-\omega) (1 +\omega +\omega^2).$$

As $1\not =\omega$, $1-\omega\not =0$, $1 +\omega +\omega^2 = 0$. Therefore,

$$1 +\omega = -\omega^2, \text{(1)}$$

$$1 +\omega^2 = -\omega. \text{(2)}$$

$$(1 -\omega +\omega^2)(1 +\omega -\omega^2) = (-\omega -\omega)(1 +\omega -\omega^2) \text{\# by (2)}$$

$$= (-\omega -\omega)(-\omega^2 -\omega^2) \text{\# by (1)}$$

$$= (-2)\omega (-2)\omega^2$$

$$= 4\omega^3 = 4.$$