Turgeon already mentioned Cardano; here's how to apply it to your polynomial.
The first thing you need to do is to depress your cubic. "Depression" is the generalization of the usual "completing the square" done on quadratics. In particular, we make a variable substitution that removes the quadratic term.
In this case, from Vieta's formulas, we know that the mean of the roots is $6/3=2$, so we let $x=u+2$. This yields the polynomial $u^3-9u-20$.
Now we come to Cardano's piece of trickery. Cardano assumes that a root $u$ of $u^3-9u-20$ can be written in the form $u=r+s$. If we make that substitution, we obtain
$(r+s)^3-9(r+s)-20=r^3+s^3+3(r+s)(rs-3)-20$
From this, we obtain the system of equations
$\begin{align*} r^3+s^3&=20\\ r^3 s^3&=27 \end{align*}$
Using Vieta's formulas again, we obtain the "quadratic" equation
$(r^3)^2-20(r^3)+27=0$
You should now be able to obtain $r$ and $s$. You already know that $x=r+s+2$ is one root of your cubic; divide that out of the original cubic, and solve the remaining quadratic. Done.