Real roots of $f(x)=8x^3-6x+1$ given in complex form

Question

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?

See my method in https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0 — lab bhattacharjee, Apr 23 '21 at 20:01
@lonestudent: ironically, this makes the resolution more difficult ! — , Apr 23 '21 at 20:49
@YvesDaoust You are right! :) When I look at your answer, the original equation actually tells us "solve me in a trigonometric way" :) — lone student, Apr 23 '21 at 21:18

score 4 · Answer 1 · answered Apr 23 '21 at 20:02

4

This phenomenon is called Casus irreducibilis. You cannot express the real roots in terms of real-valued radicals; however, a trigonometric expression is possible.

answered Apr 23 '21 at 20:02

heropup

143,828

Yes, this is a well-known fact. – lone student Apr 23 '21 at 20:06
1

I did not know this well-known fact ("I know that I know nothing") – Ataulfo Apr 23 '21 at 20:16

score 3 · Answer 2 · 2021-04-23T21:22:16.197

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.

Real roots of $f(x)=8x^3-6x+1$ given in complex form

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