Real roots of $f(x)=8x^3-6x+1$ given in complex form The function $f (x) = 8x^3-6x + 1$ has three real roots as it is easy to check (a quick way is to see the graph of f). However, Wolfram gives  the following exact values as roots:

There is no Wolfram error here because if asked for approximate root values, it gives the real numbers $-0.93969,0.17365$ and $0.76604$.
In sum, each of the three exact values when simplified must give a real number. I have tried to find at least one of these three values but it was not possible in a first attempt. Can someone describe a method to do it?

The problem is in fact very easy to solve (it was probably arranged to be so). Rewrite as
$$4x^3-3x=-\frac12$$
and compare to the trigonometric formula
$$\cos3\theta=4\cos^3\theta-3\cos\theta.$$

$$3\theta=\frac{2\pi}3+2k\pi,k=0,1,2,\\x=\cos\theta.$$

This is a case of angle trisection.

Note that by a simple rescaling of the variable, you can generalize to all equations of the form $$px^3+qx+r=0$$ that have three solutions.
