Deriving the Cubic Formula Part I: Depressing the Cubic [communicating in math]

I spent some time trying to come to grips with the derivation of the cubic formula. I was surprised to discover that all the algebra involved is elementary — at the most, a student who succeeded in Alegbra II should be able to understand the steps involved (leaving aside the imaginary numbers when it comes to actually finding the roots — but oh well, Cardano himself didn't understand them), but most of the procedures are unintuitive; it would be unclear to an observer just why we take this step or that substitution.

I'll try to hit on all the major concepts, and lay out the derivation in such a way that (hopefully) it can be followed by someone who doesn't understand the process yet.

The first thing to know is that solving a cubic is practically two separate problems. First (and this is the easier of the two steps), we use some clever factoring to "respell" our cubic so it has no x² term. When we do this we are said to be depressing the cubic. The reason this is a desirable thing to do will become clear in the next blog post; for now, just assume that our end goal is some equation of the form

y = x³ + cx + d

We'll begin, however, with a cubic equation of the form,

y = x³ + bx² + cx + d

If the x³ term has a coefficient other than 1, say, a, it's much simpler to divide the entire polynomial by a and then take that result as your cubic to be solved.

Now, the strategy for eliminating that bx² term relies on "completing the cube." One thing to which we'll recur multiple times is the Binomial Theorem, especially the expansion of a cubed binomial. Just to put it out there in eyesight,

(x + b)³ = x³ + 3bx² + 3b²x +b³

Better still for our purposes,

(x + b/3)³ = x³ + bx² + b²x/3 +b³/27

In other words, we can dispose of that bx² term by folding it into a cubed binomial. To complete the cube, we add and subtract b²x/3 +b³/27 from the cubic and follow through as below. The substitution with z at the end is just to better showcase the underlying nature of the depressed cubic with which we're left.

Note that there is also a way to depress the cubic by beginning with a substitution. We can let x = (z – b/3) and simplify, which makes the algebra easier, but, in my opinion, seems to come out of nowhere and obscures the underlying logic of what is going on. Here it is though:

Either way, we have our depressed cubic and are ready to proceed to Part II.