How to solve a cyclic quintic in radicals? Galois

How to solve a cyclic quintic in radicals?
Galois theory tells us that
${\displaystyle \frac{z^{11}-1}{z-1}=z^{10}+z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1}$ 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 ${\displaystyle {\zeta }^1,{\zeta }^2,\dots ,{\zeta }^{10}}$, 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 ${\displaystyle A_0,\dots ,A_4}$ one easily gets ${\displaystyle x_1,\dots ,x_5}$. It's easy to find ${\displaystyle A_0}$. The point is that ${\displaystyle \tau }$ takes ${\displaystyle A_j}$ to ${\displaystyle {\zeta }^{-j}{A}_j}$ and so takes ${\displaystyle A_j^5}$ to ${\displaystyle A_j^5}$. Thus ${\displaystyle A_j^5}$ can be written down in terms of rationals (if that's your starting field) and powers of ${\displaystyle \zeta }$. Alas, here is where the algebra becomes difficult. The coefficients of powers of ${\displaystyle \zeta }$ in ${\displaystyle A_1^5}$ 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 ${\displaystyle A_1}$ as a fifth root of a certain explicit complex number. Then one can express the other ${\displaystyle A_j}$ in terms of ${\displaystyle A_1}$. 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.