Deriving the Cubic Formula Part II: Solving a Depressed Cubic [communicating in math]

Although we're continuing from our work last time, we'll start afresh (i.e., variables aren't going to retain their meanings from last time) with a generic depressed cubic of the form

y = x³ + cx + d

The overarching goal here is to wind up with a quadratic equation so that we can revert to the familiar Quadratic Formula.

In this step, we're all but forced to use an early substitution. Much as I would love to accomplish the "respelling" by just factoring our starting variables (and it can be done), it's just too complicated to be bothered with (at one point, you have to complete, not the square, not the cube, but a binomial taken to the sixth power). If you're at all interested in how the algebra works, though, you can click through to this Desmos graph

https://www.desmos.com/calculator/dxik7vhi7v

where equations 17 through 9 (reading upwards) are the same function respelled over and over.

But that's really not worth it, so we'll just use a substitution. Here we're actually departing from what Cardano did in favor of a much simpler method by Vieta. We let

plug things in, and expand and simplify as below:

This last line doesn't look like a quadratic equation, but it is. Remember, we're only interested in the roots of our depressed cubic formula, which means we can set y = 0 and multiply both sides by z:

Indisputably a quadratic, so we apply the Quadratic Formula:

At this point, you might think we could plug z back into x and be done, which would make our Depressed Cubic Formula look like this:

That's mostly correct, but there is a major complication. A cubic ought to have three roots; here the plus/minus sign would seem to give us two of those, but the third one is unavailing...

Well, this isn't actually a well known fact (I didn't realize it myself until I started studying the Cubic Formula), but a square root isn't the only single-input expression with more than one output. It turns out that a cube root has three solutions, a fourth root has four solutions, and so on. Let's take 1^(1/3). Obviously, 1 is a root, but so too are –1/2 + (3^(1/2)/2)i

and its complex conjugate, –1/2 – (3^(1/2)/2)i

In fact, the three roots are perfectly spaced on an equilateral triangle inscribed in the unit circle on the complex plane.

(As it turns out, any nth roots have this remarkable property of sitting at the points of an n-gon on the complex plane.) This is fortunate, because it means that if we have only one root of z^(1/3), we can use trigonometry to find the other two.

So this, and not the plus/minus sign from the Quadratic Formula (which ends up cancelling itself out in the course of things), is what gives us three roots to each cubic.

Let's try an example. Find the zeros of

y = x³ – 15x + 4

First we'll find z as defined in the substitution above:

Then with a lot of tedious trig (not included because it's not really the point of this exercise), the three cube roots of z:

Finally, plug each in to find the three solutions for x. I'll write out the first listed just to show how the plus/minus sign ends up not mattering in the end:

I promise the other two work too, but I'm not writing out all those radicals! Those roots turn out to be 2 + 3^(1/2) and 2 – 3^(1/2).

That was an amazing amount of work to find the exact values of those roots, and remember, that's when we start with a depressed cubic. The actual cubic formula combines the steps in these two blog posts, which means just substituting one substitution into the other. It's nothing special once you've seen all the work along the way.

I'm glad to say I understand the derivation of this formula. I still think I'll stick to Newton's Method next time though.