Sum of Cube Roots of Unity/Proof 3

Theorem

Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.

Then:

$1 + \omega + \omega^2 = 0$

Proof

Observe:

\(\ds \paren {1 - \omega} \paren {1 + \omega + \omega^2}\)

\(=\)

\(\ds 1 - \omega^3\)

Difference of Two Cubes

\(\ds \)

\(=\)

\(\ds 1 - 1\)

Cube Roots of Unity

\(\ds \)

\(=\)

\(\ds 0\)

As $\omega \ne 1$, it follows:

$1 + \omega + \omega^2 = 0$

$\blacksquare$