Sum of Cube Roots of Unity/Proof 2

Proof
From Sum of Powers of Primitive Complex Roots of Unity:


 * $\displaystyle \sum_{j \mathop = 0}^{n - 1} \alpha^{j s} = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}$

Here we have that $n = 3$ and $s = 1$.

Thus $n$ is not a divisor of $s$.

Hence the result.