I'm trying to calculate the exact values of the roots of $x^3 - x^2 +1 =0$ (When I say exact value, here I mean find the roots in the form of $a + bi$ for some $a,b \in \mathbb{R}$
My first thought is to get it to a form where I can use Cardano's method and so, since 0 isn't a root, I took $x = 1/t$ and then got
$t^3 - t +1 = 0$
So by Cardano's formula, I got that
t = $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{1}{4}-\frac{1}{27}}} \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{1}{4}-\frac{1}{27}}}$
= $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}} + \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$
In $\mathbb{C}$, both $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}}$ and $\sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$ have three cube roots, and so taking $u$ and $v$ to be some particular cube roots, I know that I need to compute $\omega ^k u + \omega^l v$ for $k,l \in \{0,1,2\}$ and $\omega$ being a cube root of unity.
But I'm not sure about 1) how to compute any particular values of $u$ and $v$ in this case, as well as 2) if there is a cleaner way to approach this problem and solving monic cubic polynomials of this form in general.
Any advice would be appreciated!
Without change of variables, using Cardano method, the raw results are $$x_1=\frac{1}{3} \left(1-\sqrt[3]{\frac{2}{25-3 \sqrt{69}}}-\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}\right)$$ $$x_2=\frac{1}{3}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1-i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$ $$x_3=\frac{1}{3}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1+i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$
Now, being patient $$x_{2,3}=a\pm b i$$ where $$a=\frac{1}{3}+\frac{1}{6} \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$ $$b=\frac{\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}}{2 \sqrt{3}}-\frac{1}{2^{2/3} \sqrt{3} \sqrt[3]{25-3 \sqrt{69}}}$$