Roots of $f(x)=x^3-x^2-2x+1$

Question

We can prove using a monotony study that the function $f(x)=x^3-x^2-2x+1$ has three real roots. However, when I solve the equation $f(x)=0$ using Mathematica, I get $$x_1=\frac{1}{3}+\frac{7^{2/3}}{3 \left(\frac{1}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$ $$x_2=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$ $$x_3=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$

Since those roots are real, then maybe there's a way to write them without using the complex number $i$?

Answer 1

You need the complex numbers even when there are three real roots. That was a real puzzle when the cubic was first solved.

Read about the casus irreducibilis: https://en.wikipedia.org/wiki/Casus_irreducibilis

Answer 2

This is Example 9.2.2 in the first edition of Galois Theory by David A. Cox. He gives the polynomial, on page 239, as $y^3 + y^2 - 2 y -1.$

Easy enough to confirm the next bit with trig identities for $\cos 2t$ and $\cos 3t$ in terms of $\cos t \; .$ We have $\cos 2t = 2 \cos^2 t - 1$ and $\cos 3t = 4 \cos^3 t - 3 \cos t.$ The easiest proof is what Gauss intended, take a nontrivial seventh root of unity $\omega,$ so that $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0.$ Then take $x = \omega + \frac{1}{\omega},$ calculate $x^3 + x^2 - 2x - 1,$ and confirm that $(x^3 + x^2 - 2x - 1) \omega^3 = \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1,$ which is then zero.

The roots of $x^3 + x^2 - 2 x -1$ are $$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \; 2 \cos \left( \frac{6 \pi}{7} \right) \; . \; $$ Yours are negative of these, therefore $$ 2 \cos \left( \frac{ \pi}{7} \right) \; , \; 2 \cos \left( \frac{3 \pi}{7} \right) \; , \; 2 \cos \left( \frac{5 \pi}{7} \right) \; . \; $$

from Cox:

From Reuschle (1875)