(Revised from its original form.) Consider primes $p=6m+1$, and $p>13$:
$p=19=3\times\color{blue}{6}+1$:
Let $\beta = 2\pi/19.\,$ A root of $x^3+x^2-\color{blue}{6}x-7=0$ is $$x=2\big(\cos(2\beta)+\cos(3\beta)+\cos(5\beta)\big)$$ Note that $2+3=5$, or the last multiplier is the sum of the first two.
$p=31=3\times\color{blue}{10}+1$
Let $\beta = 2\pi/31.\,$ A root of $x^3+x^2-\color{blue}{10}x-8=0$ is $$x=2\big(\cos(2\beta)+\cos(4\beta)+\cos(8\beta)+\cos(16\beta)+\cos(30\beta)\big)$$ Similarly, $2+4+8+16=30$. Note that $\cos(30\beta) =\cos(32\beta)$.
$p=37=3\times\color{blue}{12}+1$
Let $\beta = 2\pi/37.\,$ A root of $x^3+x^2-\color{blue}{12}x+11=0$ is $$x=2\big(\cos(2\beta)+\cos(9\beta)+\cos(12\beta)+\cos(15\beta)+\cos(16\beta)+\cos(54\beta)\big)$$ Again, $2+9+12+15+16=54$.
$p=43=3\times\color{blue}{14}+1$
Let $\beta = 2\pi/43.\,$ A root of $x^3+x^2-\color{blue}{14}x+8=0$ is $$x=2\sum_{k=1}^7\cos\big(2^k\beta)$$ But alternatively, $$x=2\big(\cos(\beta)+\cos(4\beta)+\cos(11\beta)+\cos(16\beta)+\cos(21\beta)+\cos(35\beta)+\cos(88\beta)\big)$$ and, $1+4+11+16+21+35=88$.
Question: Is it true that argument multipliers of the cubic roots obey $\displaystyle\sum_{k=1}^{m-1}a_k=a_m$ for any $p=6m+1$?