Equation in radicals $x^7+7x^3-7x^2+7x+1=0$ Solve the following equation in radicals.
$$x^7+7x^3-7x^2+7x+1=0$$
It's obvious that this equation has no rational roots.
And what is the rest?   
 A: The Galois group of the polynomial is $D_{14}$, thus it is solvable.
RadiRoot is an implementation of the algorithm described in this paper:
http://www.icm.tu-bs.de/ag_algebra/software/distler/Diplom.pdf. This is the output of RadiRoot:
Let $\zeta_7$ be a primitive $7$-th root of unity and
$$\omega_1 = \sqrt[7]{ - \frac{5}{7^3}\zeta_{7}^{5} - \frac{4}{7^3}\zeta_{7}^{4} - \frac{4}{7^3}\zeta_{7}^{3} - \frac{5}{7^3}\zeta_{7}^{2} - \frac{3}{7^3}}.$$
Then a root is
$$\left(-14 \zeta _7^5+28 \zeta _7^4+28 \zeta _7^3-14 \zeta _7^2+21\right) \omega
   _1^6+\left(7 \zeta _7^5-7 \zeta _7^4-7 \zeta _7^3+7 \zeta _7^2\right) \omega
   _1^5+\left(7 \zeta _7^4+7 \zeta _7^3+7\right) \omega _1^4+\left(\zeta _7^5-2 \zeta
   _7^4-2 \zeta _7^3+\zeta _7^2-5\right) \omega _1^3+\left(\zeta _7^5-\zeta _7^4-\zeta
   _7^3+\zeta _7^2\right) \omega _1^2+\omega _1.$$
Edit.
For example, let $\zeta_7 = e^{-\frac{4 i \pi }{7}}$.
Then we have
$$\omega_1 = \sqrt[7]{\frac{-3-8 \sin \left(\frac{3 \pi }{14}\right)+10 \cos \left(\frac{\pi
   }{7}\right)}{7^3}} \approx 0.43566,$$
and a root is:
\begin{align}
\left(56 \sin \left(\frac{3 \pi }{14}\right)+28 \cos \left(\frac{\pi }{7}\right)+21\right) &\omega
   _1^6
\\-\left(14 \sin \left(\frac{3 \pi }{14}\right)+14 \cos \left(\frac{\pi }{7}\right)\right) &\omega
   _1^5
\\+\left(14 \sin \left(\frac{3 \pi }{14}\right)+7\right) &\omega _1^4
\\-\left(4 \sin \left(\frac{3 \pi }{14}\right)+2 \cos \left(\frac{\pi }{7}\right)+5\right) &\omega _1^3
\\-\left(2 \sin \left(\frac{3 \pi }{14}\right)+2 \cos \left(\frac{\pi }{7}\right)\right) &\omega _1^2
\\+\ &\omega _1 \approx -0.125215.
\end{align}
