How to solve a cyclic quintic in radicals? Galois theory tells us that $\frac{z^{11}-1}{z-1} = z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group $[2,4,8,5,10,9,7,3,6,1]$ we can partition into the five subgroups of conjugate pairs $[2,9]$,$[4,7]$,$[8,3]$,$[5,6]$,$[10,1]$.
Now put $q_1 = \zeta^2+\zeta^9$,$q_2 = \zeta^4+\zeta^7$,$q_3 = \zeta^8+\zeta^3$,$q_4 = \zeta^5+\zeta^6$,$q_5 = \zeta^{10}+\zeta^1$. So if we can solve the quintic $(q - q_1)(q - q_2)(q - q_3)(q - q_4)(q - q_5) = q^5 + q^4 - 4q^3 - 3q^2 + 3q + 1 = 0$ we would just be left to solve a few quadratic equations.
Now pari/gp tells me this quintic has the cyclic group C(5):
? polgalois(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1)
%1 = [5, 1, 1, "C(5) = 5"]

I've worked through examples of solving quadratic and cubic equations based on the galois group but when it comes to this quintic I'm completely stumped so any advice would be tremendously helpful! Thanks.
Edit: Thanks to Robin Chapman the problem is reduced significantly. Let $\omega$ be a primitive 5th root of unity (which is easy to express in radicals), it only remains to express $(q_1 + \omega q_2  + \omega^2 q_3  + \omega^3 q_4  + \omega^4 q_5)^5$ in terms of rationals and powers of $\omega$ (and then everything can be substituted back and solved easily). We know this is possible because the term is fixed by the quintics galois group, how to actually perform this evades me but I will try to find a way.

The solution of a solvable quintic,
$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$
can be given by,
$$x = \frac{1}{5}\left(-a+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$
and the $z_i$ are the roots of the quartic. However, there is also a form where one takes a fifth root only once.

$p=11$

Let,
$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0$$
The five roots $x_k$ for $k=0,1,2,3,4$ 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,
$$a=\tfrac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$
$$b=\tfrac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$
$$\beta_k = \zeta_5^k\,\left(\frac{ab}{11}\right)^{1/5}$$
$$\zeta_5 = e^{2\pi\,i/5}$$

$p=31$

As discussed in this post,
$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$
$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$
where,
$$a=\tfrac{31}{4}\left(11+5\sqrt{5}+\sqrt{-10(25-11\sqrt{5})}\right)$$
$$b=\tfrac{31}{4}\left(-1-5\sqrt{5}+\sqrt{-10(1525-\sqrt{5})}\right)$$
$$\beta_k = \zeta_5^k\,\left(\frac{ab}{31}\right)^{1/5}$$
and so on for other prime $p=10n+1$.
