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I. Quintics

For an example of a cyclic quintic, we have for $p=11$,

$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0\tag1$$

The five roots $x_k$ for $k=0,1,2,3,4$, in radicals, are,

$$x_k = \frac{-1}{5}\left(\frac{1}{\beta_k^{-1}}+\frac{1}{\beta_k^0}+\frac{11}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$

where,

$$a=\tfrac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$

$$b=\tfrac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$

$$\beta_k = \zeta_5^k\,\left(\frac{ab}{11}\right)^{1/5}$$

$$\zeta_5 = e^{2\pi\,i/5}$$

The case for $p=31$ is discussed in this post, while minimal polynomials for other prime $p=10n+1$ can be found here.

II. Septics

Q: Can the roots of the minimal polynomial of $x = 2\big(\cos\big(\tfrac{2\pi}{43}\big)+\cos\big(\tfrac{12\pi}{43}\big)+\cos\big(\tfrac{14\pi}{43}\big)\big)$,

$$x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7x - 49=0\tag2$$

be also expressed in an analogous form,

$$x_k = \frac{1}{7}\left(\frac{a_0}{\beta_k^{-1}}+\frac{\pm1}{\beta_k^0}+\frac{a_1}{\beta_k^1}+\frac{a_2}{\beta_k^2}+\frac{a_3}{\beta_k^3}+\frac{a_4}{\beta_k^4}+\frac{a_5}{\beta_k^5}\right)$$

$$\beta_k = \zeta_7^k\,\left(\frac{\gamma}{43}\right)^{1/7}$$

$$\zeta_7 = e^{2\pi\,i/7}$$

where the $a_i$ and $\gamma$ are algebraic numbers of degree at most $6$? In general, can we do this for all cyclic septics?

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Tito Piezas III
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    Undoubtedly you know this, but if we let $K$ to be the splitting field of a cyclic septic, then by basic Galois theory $L=K(\zeta_7)$ is a root extension $\Bbb{Q}(\zeta_7,u^{1/7})$ for some $u\in \Bbb{Q}(\zeta_7)$, so the numbers $x_k$ can be written in terms of $\zeta_7$ and $u^{1/7}$. I don't know exactly how far that could take us. We always have that $L$ is Galois and abelian over $\Bbb{Q}$, so by Kronecker-Weber it is a subfield of some cyclotomic field. In your case the conductor is $43\cdot7$ but it could be something else also. – Jyrki Lahtonen Jun 29 '15 at 07:04

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