TITLE: A new way of solving cubics?
QUESTION [10 upvotes]: I found this (from http://www.quora.com/Mathematics/What-are-some-interesting-lesser-known-uses-of-the-quadratic-formula):

So my question is: Can this be generalized to solve any depressed cubic [in effect, all the cubics]. Maybe not exactly this method, but some other repeated manipulation of the the quadratic formula? And finally, which is my main aim, can this generalized method give us exact, not-so-messy answers ie. respite from the Cardano's method?

REPLY [7 votes]: One straightforward way to see that this method cannot be valid for arbitrary cubics: The porcess can only produce solutions using square roots, while there are cubics that do require $\sqrt[3]~$ to express the roots (simplest case: $x^3-2=0$, but cf. also Cardano's general result).
Also geometrically, taking square roots is possible with straightedge and compasses and angle trisection is not - but would be if we could solve cubics!