So now we can talk turkey about the rest of (my version of) the proof.
1) divide through by the leading coefficient a. (get a monic univariate polynomial equation)
2) substitute x = (t - b/(3a)) and rearrange to get a new cubic equation in terms of t, which manifestly has no quadratic term (that is, the coefficient vanishes - get an MDUPE. This type of substitution is called a /Tschirnhausen substitution/. Protip: it is equally valid to derive the quadratic formula. Try it!)
3) give simple names to the coefficients of this new cubic in t, which are given in terms of the original a,b,c,d. Call them p and q, say. Bookeeping. Just never forget that you want to get back to those a,b,c,d when you're all done.
4) set t = w - p/(3w). This is a type of substitution, a further trick, which may be called a /Viete substitution/. You must convince yourself that there really exists such a w! This is an important exercise which I leave for you (use the quadratic formula).
5) If you're good that this "w" really exists, then you can yet-agian substitute in the above, in place of t.
6) Rearrange the above. You get a polynomial of degree six in terms of w, with just one other term of degree three in terms of w, and a constant term (with respect to w). note that the above p and q terms act in the role of coefficients just now. In other words... you have a quadratic in terms of w^3, so we may as well call w^3 = z.
7) Solve for z using the quadratic formula. Since z = w^3, simply (and brashly) solve for w, by just... taking the third root!
8) At this point, I found it helpful to identify and specify, for my personal use, a number of "discriminant" forms, and "conjugate" forms. It is left to you as an exercise to know exactly what those things are, to keep your bookkeeping straight.
9) This is the big non-obvious trick which Derbyshire invoked, and which I used to complete my own proof:
[math] (w + \bar{w})^3 + p(w + \bar{w}) + q = 0 [/math]
cont.