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Let $f=X^3+X^2-2X-1$ be a polynomial with the three roots $x_1,x_2,x_3$ with $x_1=2\text{cos}(\frac{2 \pi}{7})$. We define $z:=(x_1-x_2)(x_1-x_3)(x_2-x_3)$.

I want to find a radical expression for $x_1=2\text{cos}(\frac{2 \pi}{7})$ over $\mathbb{Q}[z]$.

I found some information here and here and with a computer program I calculated that $$x_1=\frac{1}{3}\left(-1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} (1+3 i \sqrt{3})}}+\sqrt[3]{\frac{7}{2} (1+3 i \sqrt{3})}\right).$$

I think this expression is not the one I am searching for but I wonder how I can find it by calculating it in a correct way (without using a program).

Finally, what's the importance of $z$ and $\mathbb{Q}[z]$? Is there a way to calculate $z$? Using a computer program I found out that $z \in \mathbb{Q}$, which confuses me.

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I wonder how I can find it by calculating it in a correct way (without using a program).

The cubic formula has been around for the last $500$ years. Feel free to use it.


I think this expression is not the one I am searching for.

Why ? Because it contains in its expression the square root of $-1$, despite being obviously a real number, and not an imaginary one ? If so, then please see casus irreducibilis.


Using a computer program I found out that $z\in\mathbb Q$, which confuses me.

In that case, read Niven's theorem, and let the confusion dissipate.

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  • $\begingroup$ Thank you, I'll check your links. I think I am not allowed to use the cubic formula. Anyway, what is the cheapest way to calculate $z$? It looks to be a bit expensive. $\endgroup$ – user152546236 May 24 '14 at 8:50
  • $\begingroup$ Are you asking me how to numerically compute by hand the value of $x_1$ up to a certain number of decimal places with the least effort, by using the fact that it is a root of a given cubic equation ? If so, then there are plenty of numerical methods which you might employ to that end. $\endgroup$ – Lucian May 24 '14 at 9:13
  • $\begingroup$ No, I want to do it per hand-calculating. I think the best method is using the complex forms. $\endgroup$ – user152546236 May 24 '14 at 9:14
  • $\begingroup$ I'm afraid I don't understand what you mean by hand calculating. $\endgroup$ – Lucian May 24 '14 at 9:23
  • $\begingroup$ Without using a computer. $\endgroup$ – user152546236 May 24 '14 at 9:26
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Expanding the complex expressions, the three roots are given by$$x_1=\frac{2}{3} \sqrt{7} \cos \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)-\frac{1}{3}$$ $$x_2=-\frac{1}{3}-\sqrt{\frac{7}{3}} \sin \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)-\frac{1}{3} \sqrt{7} \cos \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)$$ $$x_3=-\frac{1}{3}+\sqrt{\frac{7}{3}} \sin \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)-\frac{1}{3} \sqrt{7} \cos \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)$$ and, effectively, $x_1=2 \cos \frac{2 \pi} {7}$. I suppose that you see the common stucture which appears in the roots.

Added later to this answer

Brute force gives for the roots $$x_1=\frac{1}{3} \left(-1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}\right)$$ $$x_2=-\frac{1}{3}-\frac{7^{2/3} \left(1+i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}$$ $$x_3=-\frac{1}{3}-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}$$

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    $\begingroup$ I thought in a radical expression there shouldn't be any $\text{sin},\text{cos}$ or $\text{tan}$ functions. And how can I see that $z \in \mathbb{Q}$? $\endgroup$ – user152546236 May 23 '14 at 21:58
  • $\begingroup$ How can I find the complex expressions of $x_2$ and $x_3$? $\endgroup$ – user152546236 May 24 '14 at 8:45
  • $\begingroup$ I shall add these awful expressions in my answer. $\endgroup$ – Claude Leibovici May 24 '14 at 8:47
  • $\begingroup$ The problem is that only $x_1$ is given. So I don't know how to find $x_2$ and $x_3$ by hand-calculating; as you said - awful expressions. $\endgroup$ – user152546236 May 24 '14 at 8:51
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    $\begingroup$ Yes ! Vieta for sure. This is the only way. $\endgroup$ – Claude Leibovici May 24 '14 at 9:16

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