If $p$ is a prime number and $ a_0,a_1,\ldots , a_{p - 1} $ are rational numbers satisfying $$ a_0 + a_1 \alpha + a_2 \alpha ^2 + \ldots + a_{ p - 1} \alpha ^{ p - 1} = 0 $$ where $$ \alpha = \cos(\frac{2 \pi}{p}) + i \cdot \sin(\frac{2 \pi}{p} ) = e^{\frac{2i\pi}{p}} $$ then $a_0 = a_1 = \ldots = a_{p -1}$.
I think that it is helpful to consider:
(i) $ g(x) = 1+ X^1 + \ldots + X^{p - 1}$ roots(particularly, $\alpha$ and its conjugate) and the fact that it is irreducible.
(ii) $f(x) = a_0+ a_1X^1 + \ldots + a_{p - 1}X^{p - 1}$.
Hint:
If $p$ is prime, $1+X+\dots+X^{p-1}$ is the $p$-th cyclotomic polynomial and it is irreducible. Actually it is the minimal polynomial of $\alpha$ and its conjugates.
What can you deduce for $f(X)$?