Solving cubics - Cardano's Method

With apologies for the lack of updates, now for something completely different!

Below is a general method of tackling cubics. I have seen the method explained in many an account, but conciseness and clarity are often amiss (if ironically, in the mistaken hope that the needless additions and illustrations help further understanding!).

This is interesting not for actual computational use (though perhaps some will like that), but for the idea behind it. Simply put we reduce what we need to a solving a quadratic and then extract cube roots at the end.

So suppose our cubic is of the form,
ax^3 + bx^2 + cx + d = 0
First off we use the substitution y = x - b/(3a) to reduce this to what's known as a 'depressed cubic' of the form,
y^3 = 3Gy + H

Now recall the expansion,
(p+q)^3 = 3pq(p+q) + p^3 + q^3

We note on comparison then that y=p+q is a solution to our cubic if we take,
pq = G and p^3 + q^3 = H

Now form the quadratic with p^3 and q^3 as roots, that is,

t^2 - Ht + G^3

Solving this we get two values which we can arbitrarily call p^3 and q^3 respectively.
Now we take cube roots for each, and in each case we get three cube roots (call the cube roots of unity 1, w and w^2 say, where w = (-1+i.sqrt(3))/2 or (-1-i.sqrt(3))/2, which you can verify for yourself).

Since y=p+q we need be careful which pairs to match up, but on recalling that pq=G we get,
p+q, wp+(w^2)q and (w^2)p+wq

as the three desired cube roots!