On the trigonometric roots of a cubic We have,
$$x^3+x^2-2x-1=0,\quad\quad x =\sum_{k=1}^{2}\,\exp\Bigl(\tfrac{2\pi\, i\, (6^k)}{7}\Bigr)\\
x^3+x^2-4x+1=0,\quad\quad x =\sum_{k=1}^{4}\,\exp\Bigl(\tfrac{2\pi\, i\, (5^k)}{13}\Bigr)\\
x^3+x^2-6x-7=0,\quad\quad x =\sum_{k=1}^{6}\,\exp\Bigl(\tfrac{2\pi\, i\, (8^k)}{19}\Bigr)$$
and so on. It seems the general formula is, given a prime $p=6m+1$ and the unique solution $u,v$ to,
$$u^2+27v^2=4p$$
then,

$$x^3+x^2-2mx+\frac{1-3p\pm p u}{27}=0\tag1$$

for the appropriate sign $\pm$ and where $x =\sum_{k=1}^{2m}\,\exp\Bigl(\tfrac{2\pi\, i\, (\color{blue}c^k)}{p}\Bigr)$ and appropriate integer $\color{blue}c$. For example, for $p=7$, we have
$$(-1)^2+27\times1^2=4\times7$$
and using $u=-1$ on formula $(1)$, we recover the first example.

Q: Is it possible to find a formula for the sign $\pm$ and/or $\color{blue}c\,$ as a function of $p$?

$\color{blue}{Update}$: It seems the answer to the part about the sign is yes. While checking the OEIS for the sequence of solutions $u=1, 5, 7, 4, 11, 8,\dots$, it turns out there is signed version using the Kronecker symbol(?). Thus,

$$x^3+x^2-2mx+\frac{1-3p- p\,u'}{27}=0\tag2$$

where $u'=1, -5, 7, 4, -11, -8,\dots$ and given by A123489.
 A: If $p = 3n+1$ and $u,v$ satisfy $u^2+27v^2 = 4p$ with $u \equiv 1 \pmod 3$
then taking this relation modulo $27$ we get $p \equiv 7u^2$, and then
$1-3p-pu \equiv 1-21u^2-7u^3 \equiv 7(1-u)^3 - 12(1-u)^2 + 9(1-u) \equiv 0$
because $1-u \equiv 0 \pmod 3$.
Then if we call $w = \frac 1 {27} (1-3p-pu)$, $w$ is an integer.
If $p$ is odd then $w = u = v \pmod 2$ and $n$ is even.
The cubic $x^3+x^2-n+w$ has discriminant 
$D = n^2+4n^3-4w-27w^2-18nw$.
$27D = 3(p-1)^2+4(p-1)^3-4(1-3p-pu)-(1-3p-pu)^2-6(p-1)(1-3p-pu)$
which after simplification gives $27D = p^2(4p-u^2) = 27p^2v^2$ and so $D = p^2v^2$
Since this is a square, the Galois group is cyclic and 
after computing the interpolating polynomials (or by factoring the cubic and solving the remaining quadratic), we get that if $x$ is a root then the other two are
$\frac 1v(x^2 + x(1+((n+9w)/p-v)/2) - (n+((n^2-3w)/p+v)/2))$
and its conjugate obtained by switching the sign of $v$.
Except the division by $v$ at the front, the divisions inside those coefficients all result in integers because of parity considerations, and the equations $27w \equiv -3n  \equiv 1 \pmod p$.
Therefore the lattice $\Bbb Z[x_1] = \langle 1,x_1,x_1^2 \rangle$ has index $|v|$ in the lattice $\Bbb Z[x_1,x_2,x_3] = \langle 1,x_1,x_2 \rangle = \langle x_1, x_2, x_3\rangle$, and so the discriminant of this lattice is $p^2$. Since there is no unramified extension of $\Bbb Q$, it can't get any smaller and so $\Bbb Z[x_1,x_2,x_3]$ is its ring of integers and we have an integral normal basis.
Now this extension must be the class group of conductor $p$ corresponding to the unique subgroup $H$ of index $3$ in $G = (\Bbb Z/p \Bbb Z)^*$, which is trigonometric by the Kronecker Weber theorem.
Then we have another integral normal basis for the ring of integers, $\{s_K = \sum_{k \in K} \zeta_p^k \mid K \in G/H \}$
Looking at how the Galois group can act on those elements shows that we must have $x_1 = \pm s_K$ for some $K$. Finally, since $x_1+x_2+x_3 = -1 = \sum s_K$, we have $x_1 = s_K$ for some $K \in G/H$
