Given,
$$p=u^2+27v^2=6m+1\tag1$$
and the cubic,
$$x^3+x^2-2mx+N=0\tag2$$
with its constant expressed in terms of $(1)$ as,
$$N = \frac{1}{27}(1-3p\pm2pu)\tag3$$
and the sign $\pm u$ chosen appropriately. (The discriminant is $D=-108p^2v^2$, so all real roots.) Starting with Ramanujan's general cubic identity, they are a special case, yielding the simple,
$$(a+b\,x_1)^{1/3}+(a+b\,x_2)^{1/3}+(a+b\,x_3)^{1/3}=\big(c+\sqrt[3]{dp}\big)^{1/3}\tag4$$
for some rational $a,b,c,d$.
Question: Is it true that if $p=u^2+27v^2=6m+1$ is prime, then a root of $(2)$ is always a sum of the $p$th root of unity of form,
$$x = \sum_{n=1}^{2m}\,\exp\Bigl(\frac{2\pi\, i\, k^n}{p}\Bigr)$$
for some integer $k$?
Example: We have $p=127=10^2+27\cdot1^2=6\cdot21+1$, so
$$x^3+x^2-(2\times21)x+80=0,\quad\quad x =\sum_{n=1}^{2\times21}\,\exp\Bigl(\frac{2\pi\, i\, (5^n)}{127}\Bigr)$$
with all three roots $x_i$ yielding,
$$(-2+x_1)^{1/3}+(-2+x_2)^{1/3}+(-2+x_3)^{1/3}=-\big(19-3\,\sqrt[3]{254}\big)^{1/3} = -0.097378\dots$$
P.S. This generalizes Question 3 of this post, and is also related to this post.
