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.
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 $$x_1=\frac{1}{3} \left(-1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}\right)$$ $$x_2=-\frac{1}{3}-\frac{7^{2/3} \left(1+i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}$$ $$x_3=-\frac{1}{3}-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(1+3 i \sqrt{3}\right)}$$