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 $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, 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].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.