The sum $\sum_{t \not \mod - 1} \chi[t(t + 1)]$

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It's from Multiplicative Number Theory, Davenport, pp. 24.

In calculating $\tau^{2}$ where \[ \tau = \sum_{x = 1}^{q - 1} \chi(n)e_{q}(x), \] $\chi(n)$ Dirichlet character mod 3 and $e_{q}(x) = e^{2 \pi i x/ q}$, I see a step which I can't understand.

It is easy to see that \[ \sum_{t} \chi(t(1 + t)) = A + B\omega + C \omega^{2}. \] But how can we verify that \[ \sum_{t} \chi(t(1 + t)) = A + B\omega, \] where $\omega$ is a cubic root of unity?

Wouldn't cubic roots of unity be linearly independent of one another?