The following equation is solvable by radicals. How to get all the roots in terms of radicals?
$ x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = 0$
The transcendental solution is given by
$ \cos(k\pi/13)+i\sin(k\pi/13)$ for $ k = 1, 2,\dots.,12$. Only a solution by radicals is acepted as an answer.
Let $\zeta = e^{2\pi i / 13}$.
$(\mathbb{Z}/13\mathbb{Z})^{\times}$ is cyclic of order $12$. You get something cubic over $\mathbb{Q}$ by adding $\zeta^k$ where $k$ runs through the subgroup of index $3$ or its cosets:
$$A := \zeta^1 + \zeta^5 + \zeta^8 + \zeta^{12}, \; B := \zeta^2 + \zeta^3 + \zeta^{10} + \zeta^{11}, \; C := \zeta^4 + \zeta^6 + \zeta^7 + \zeta^9$$ all have minimal polynomial $X^3 + X^2 - 4X + 1$, so they can be expressed by radicals by the cubic formula.
The next step is to consider sums of $\zeta^k$ where $k$ runs through the subgroup of index $6$ and cosets:
$$a = \zeta^1 + \zeta^{12}, \; b = \zeta^2 + \zeta^{11}, \; c = \zeta^3 + \zeta^{10},$$ $$d= \zeta^4 + \zeta^9, \; e = \zeta^5 + \zeta^8, \; f = \zeta^6 + \zeta^7.$$
These are determined by the equations $$a + e = A, \; ae = C, \; b+c = B, \; bc = A, \; d+f = C, \; df = B$$ which are quadratic equations over $\mathbb{Q}(A,B,C)$. For example, it follows that $$a = \frac{A \pm \sqrt{A^2 - 4C}}{2}.$$
Finally, once you know that $\zeta + \zeta^{12} = a$ and $\zeta \cdot \zeta^{12} = 1,$ you can solve $$\zeta = \frac{a \pm \sqrt{a^2 - 4}}{2} = \frac{a}{2} + i \sqrt{1 - a^2/4}.$$