Roots of $8x^3-4x^2-4x+1$

Question

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 $$\cos\frac{\pi}{7} = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i \sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$

Answer 1

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.

Answer 2

First, with regard to the appearance of $i$ in the solutions, it's very well known that, when an irreducible cubic has three real roots, then any expression of those roots in terms of radicals must involve the imaginary unit $i$. This is the so called casus irreducibilis and was perhaps the most fundamental historical motivation for the acceptance of complex numbers.

Second, with regard to using Mathematica to see that these expressions are the same, you can use the RootReduce command to put either of these into the same canonical form. Thus, for example, the following input yields zero:

RootReduce[(1+7^(2/3)/((-1+(3*I)*Sqrt[3])/2)^(1/3)+
  ((7*(-1+(3*I)*Sqrt[3]))/2)^(1/3))/6 - Cos[Pi/7]]

This works in WolframAlpha as well.