You could also try guessing some obvious solutions and factor them, which will simplify the cubic to a quadratic. But I didn't see any just by glancing at the equation.
The trigonometric form is more simple.
Any cubic can be put into the form.
$$t^3+pt+q=0$$
which has solution
$$t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right) \quad \text{for} \quad k=0,1,2$$
see here http://en.wikipedia.org/wiki/Cubic_function#Trigonometric_.28and_hyperbolic.29_method