Find the roots of $x^3 -6x^2 +13x -12$ I am trying to find the roots of $$\tag{1} x^3 -6x^2 +13x -12$$ by applying the method outlined here (I think this Cardan’s method).
Letting $y= x-2$, we can transform $(1)$ into $$\tag{2} y^3 + y -2=0.$$
It is clear that $1$ is a root of $(2)$, and, from the formula $$u - \frac{h}{u}=1,$$ we get $$\tag{3} u= \frac{3+ \sqrt{21}}{6}.$$ 
Given that the formula for the other two roots of $(2)$ are $$u\omega -\frac{h\omega^2}{u}, \quad u\omega^2 -\frac{h\omega}{u},$$ we can use $(3)$ and the fact that $\omega=\frac{1}{2}\Big(-1+\sqrt{3}i \Big)$ in order to find them, but this seems an incredibly tedious method. 

Question: Is there a more expedient way to evaluate the last two roots of $(2)$?

Note: I am aware that you can use the fact that $x=3$ is a root of $(1)$ in order to factorise the equation, but I am trying here to use the Cubic Formula.

$$x^3+x-2= 0$$
We can eliminate the $x^1$ by using a quadratic tschirnhausen transformation $y= x^2+mx+n$, to get the cubic in a binomial form
$$(y-(x_{1}^{2}+mx_{1}+n))(y-(x_{2}^{2}+mx_{2}+n))(y-(x_{3}^{2}+mx_{3}+n)) = 0$$
Collect the new variable $y$
$$y^3+(\dots \dots)y^2+(\dots \dots)y+(\dots \dots) = 0$$
Solving for the unknowns $m$, $n$ to eliminate the $y^2$ and $y$ term
$$y^3+(2-3n)y^2+(1-6m+m^2-4n+3n^2)y+(-4-2m-2m^3-n+6mn-m^2n+2n^2-n^3) = 0$$
Therefore
$n= \frac{2}{3}$, $m = \frac{9\pm 2\sqrt{21}}{3}$
Using the negative case, the binomial cubic becomes
$$y^3-\frac{112 \cdot {112-6\sqrt{3\cdot 112}}}{54} = 0$$
$$y^3 = (\frac{2}{3})^3\cdot 7\cdot (112-24\sqrt{21})$$
$$y = \frac{2}{3}\sqrt[3]{7(112-24\sqrt{21})}$$
Don't forget $x^2+mx+n-y = 0$ and check for sign
$$x = -\frac{3}{2}+\frac{\sqrt{21}}{3} \pm \frac{1}{2}\sqrt{\frac{47}{3}-4\sqrt{21}+\frac{8}{3}\sqrt[3]{7(112-24\sqrt{21})}}$$
The trigonometric solution is
$$x = 2\sqrt{\frac{-1}{3}} \cos(\frac{1}{3}\arccos(-\sqrt{-27}))$$
EDITED to adjust the type settings
