Solve the equation $x^3-6x-6=0$ 
Evaluate the roots of
$$x^3-6x-6=0$$

I solved it using Cardano's method, but I'm looking for other elementary approaches through substitutions and properties of polynomials.
Thanks.

$\text{Let } x = z + \frac{2}{z}\\\begin{align*}\left(z+\frac{2}{z}\right)^3 - 6\left(z+\frac{2}{z}\right) - 6 &= 0\\z^6 - 6z^3 + 8 &= 0\\z^3 &= 3\pm1\, (\text{using the quadratic formula})\\z^3 &= 4, 2,\\z &= \sqrt[3]{2},\sqrt[3]{4}\end{align*}$
Now just substitute back one value for $z$ to get one root:
$x = \sqrt[3]{2} + 2\sqrt[3]{4}$
The rest of the roots can now be gotten through long division, or by using the imaginary cube roots of $2$ or $4$. 
