<s>[INST] Roots of $8x^3-4x^2-4x+1$ It is known that the roots of polynomial $8x^3-4x^2-4x+1$ are $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$ and $\cos\frac{5\pi}{7}$.
However this is what Wolfram Alpha/Wolfram Mathematica gives:
$$x = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
   \sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
   \sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i
   \sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i
   \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i
   \sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1+i
   \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
These must be the same roots, but is there a way to show it? Or at least a way to show that these roots given by Wolfram are real?
Can you show that 
$$\cos\frac{\pi}{7}  = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
   \sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
   \sqrt{3}\right)}$$
 [/INST] One can show for example, that by putting $a= 2 \cos(\pi/7)$, and $b=a^2-1$, that if $ab=a+b$ and $b^2=1+a+b=1+ab$, then you are indeed dealing with the chords of the heptagon, and that these values are correct.  
It is relatively easy to show by using coins on a table, that these two relations correspond to the solutions that $a$ and $1/b$ solve $x^3 - x^2 - 2x + 1$, in which case $a/2$ solves the equation in the OP. 
When i saw the expansion of the root into complex numbers, what immediately sprang to mind was the construction of the heptagon using a trisector, given in page 199 of Conway + Guy 'The book of Numbers'.  The basic construction involves a hexagonal lattice (Eisenstein integers), with a circle of radius 2 centred on some point, this becomes the circum-circle.
One constructs the x-axis through the centre, and a 'y-axis', passing through the first point going through the point U at cis(120 degrees)(ie $(-1+\sqrt{-3})/2$.  
The point V is an other lattice point, at $-3 \sqrt{-3})$.  The line through U and V intersects the 'y-axis'.  This minimal angle is trisected, the cut nearer the 'y-axis' is extended to the x-axis, as are the two more lines, that make 60-degrees with the first.  
These strike the x-axis, and perpendiculars from these three points strike the vertex-curve at the vertices of the heptagon.  
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