Computing the Legendre symbol of -3, $(\frac{-3}p)$ I'm working on Ireland and Rosen, exercise 6.8.

Let $\omega=e^{2\pi i/3}$ satisfying $\omega^3-1=0$. Show that $(2\omega-1)^2=-3$ and use this result to determine $(\frac{-3}p)$ for $p$ an odd prime.

I've already found that $0=\omega^3-1=(\omega-1)(\omega^2+\omega+1)$ so since $\omega\ne1$, $\omega^2+\omega+1=0$ and so computing
$$
(2\omega+1)^2=4\omega^2+4\omega+1=4(\omega^2+\omega+1)-3=-3
$$
Now putting $\tau=2\omega+1$, I've found that for any odd prime,
$$
\tau^{p-1}=(\tau^2)^{(p-1)/2}=\left(\frac{-3}p\right)
$$
by property of the Legendre symbol, so $\tau^p=(\frac{-3}p)\tau$.
Next, I should find another way to compute $\tau^p$ to equate $(\frac{-3}p)\tau$ with something else. I may need the result that
$$\tau^p=(2\omega+1)^p=(2\omega)^p+1\pmod p$$
which should take different values according some condition on $p$.
Any hint?

answer: use $\omega^3=1$ and $2^p=2\bmod p$.
