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

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
