Solving the sextic $t^6-t^5+t^4-t^3+t^2-t+1$ Ian Stewart has this problem in his Galois theory textbook.

Solve the sextic equation
  $$t^6-t^5+t^4-t^3+t^2-t+1=0$$
  satisfied by a primitive 14th root of unity in terms of radicals.

If I let $\zeta$ be this primitive root, then $\zeta$ is clearly radical as $\zeta^{14}=1$. Then aren't the solutions just $\zeta$, $\zeta^{3}$, $\zeta^{5}$, $\zeta^{9}$, $\zeta^{11}$, and $\zeta^{13}$?
All of these expressions are radical.
However Ian Stewart gives the hint: set $u=t+\frac{1}{t}$. How does this help, and is what I've done incorrect?
Thanks

What you have done is entirely correct, but the point of Stewart's hint is that the 14th root of unity does not need seventh roots to express ($\zeta=\sqrt[7]{-1}$); it only needs square and cube roots. Divide the equation by $t^3$:
$$t^3-t^2+t-1+\frac1t-\frac1{t^2}+\frac1{t^3}=0$$
We have
$$u=t+\frac1t$$
$$u^2=t^2+2+\frac1{t^2}$$
$$u^3=t^3+3t+\frac3t+\frac1{t^3}$$
$$u^3-u^2-2u+1=0$$
and the last equation is the minimal polynomial of $2\cos\frac{2\pi}{14}$. Let $C=\sqrt[3]{\frac{7+7\sqrt{27}i}2}$ and $\omega$ a primitive third root of unity, then
$$u=\frac13\left(1-\omega^kC-\frac7{\omega^kC}\right),\ k\in\{0,1,2\}$$
and the roots of the original polynomial may be recovered as $t=\frac{u\pm\sqrt{u^2-4}}2$.
Among other things, this shows that the heptagon and 14-gon are constructible with the help of an angle trisector.
