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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

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    $\begingroup$ set u=t+1/t Hint: divide by $t^3$ and arrange as to get a cubic in $u$. $\endgroup$ – dxiv Oct 5 '16 at 4:30
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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.

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  • $\begingroup$ I'm interested in geometry, but don't know much about Galois theory, although I've seen the basic notions before in an algebra class. Can you elaborate on the last sentence just a little bit? $\endgroup$ – Alfred Yerger Oct 5 '16 at 5:52
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    $\begingroup$ @AlfredYerger Normal compass-and-straightedge constructions can only solve quadratic equations. With a trisector you can also solve any cubic that has all its roots real; by reducing the sextic in $t$ to the cubic in $u$ which does have all real roots, we show that $2\cos\frac{2\pi}{14}$ - and thus the heptagon and 14-gon, since their internal angles are multiples of $\frac{2\pi}{14}$ - is constructible with a trisector. $\endgroup$ – Parcly Taxel Oct 5 '16 at 11:26

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