<s>[INST] Finding all the complex roots of $x^3 - x^2 + 1 = 0$ 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!
 [/INST] 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
$$a=\frac{1}{3}+\frac{1}{6} \sqrt[3]{\frac{1}{2} \left(25-3
   \sqrt{69}\right)}+\frac{1}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$
$$b=\frac{\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}}{2
   \sqrt{3}}-\frac{1}{2^{2/3} \sqrt{3} \sqrt[3]{25-3 \sqrt{69}}}$$
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