Solving the Cubic

Introduction

I've always thought that the explanations I've read of Cardano's solution to the general cubic were unnecessarily complicated.

Whenever I ask someone who has been exposed to the method to explain it, the usual answer is that they don't remember how it went. This is my attempt to explain it in such a way that it has a better chance of being remembered. If nothing else, I'll remember it better!

Since most approaches get very general very quickly, I'll do something else instead. I'll start with a progression of carefully chosen concrete examples. Then I'll generalize. The examples I'm going to use are:

x³ - 3x + 2 = 0
x³ - 3x = 0
x³ - 3x + 1 = 0

Along the way I'll look at the quadratic equation and completing the square, describing that technique in a way that connects with solving the cubic.

The first example won't need complex numbers, but soon after they'll be heavily used. If you're unfamiliar with complex numbers you're welcome to try and read this essay, but it will soon become really hard to follow. You've been warned!

An Example

The example I like to start off with is x³ - 3x + 2 = 0.

Simple Solution

The standard way to solve this is to try a few simple values: 1, -1, 2, -2. It turns out that 1 is a solution! That means (x-1) is a factor of x³ - 3x + 2 = 0. Polynomial division gives that

x³-3x+2 can be factored as (x-1)(x² + x - 2).

This tells us that the solutions of x³ - 3x + 2 = 0 are given by the solutions of x-1=0 and x²+x-2=0.

Solving for the quadratic factor can be done in many ways. In this case there is an obvious factorization x² + x - 2 = (x-1)(x+2). So, x³ - 3x + 2 = (x-1)²(x+2).

The roots are -2 and 1 (with multiplicity 2).

The only problem with this solution technique is that it depends on somehow intuiting a linear factor. When you can do it, you're in business. When you can't, you're stuck. Cardano's technique is more work, but it will always get you the answer.

Creative Solution

Here's how this example is solved using Cardano's technique.

The crucial creative step (in my opinion) is to replace x with (w+1/w), and to solve for w. This leads to a sixth order equation in w, BUT, it's a sixth order equation that's easy to solve. Here are the details:

x³ - 3 x + 2 =
(w+1/w)³ - 3(w+1/w) + 2 =
w³ + 3 w² (1/w) + 3 w (1/w)² + (1/w)³ - 3 (w+1/w) + 2 =
w³ + 3 w + 3 (1/w) + (1/w³) - 3 (w+1/w) + 2 =
w³ + (1/w³) + 2.

So x³-3x+2=0 becomes w³+1/w³+2 = 0

This is a sixth order equation in w. The trick to solving it is to realize that it's really a quadratic in w³. Replacing w³ with z, we get:

z + 1/z + 2 = 0

ie z² + 1 + 2z = 0.

Solving the quadratic using whatever technique you like (I like z² + 1 + 2z = (z+1)² ) you get z = -1.

So, w³ = -1.

A solution to this is w = -1, which gives the solution x = w + 1/w = -1+1/-1 = -2.

Polynomial division now gives

x³-3x+2 can be factored as (x+2)(x² - 2x + 1).

x² - 2x + 1 = (x-1)², so we have that the solutions are x = -2 and 1 (with multiplicity 2).

If you know something about complex numbers, you might have noticed that I could have used other solutions to w³ = -1. The three solutions are:

w = -1, 1/2 + i √3/2, 1/2 - i √3/2

Choosing w = 1/2 + i √3/2, we get the solution

x = w + 1/w = (1/2 + i √3/2) + (1/2 - i √3/2) = (1/2)+(1/2) = 1

Something to notice here is that we managed to get a real solution to the cubic by using complex numbers!

Also, in this particular example, using complex numbers is optional. By choosing the w= -1 solution to w³= -1 we successfully avoided dealing with them. This doesn't always happen! When I wrote that this example was carefully chosen, I meant it.

Another Example

To warm up a little more, consider x³ - 3x = 0.

Really Simple Solution

This is very easy to solve without Cardano's technique. Since there's no constant term, x is a factor.

x³-3x can be factored as x(x² - 3).

This tells us that the solutions are x = 0, √3, -√3.

Using The Sledgehammer

Replace x with (w+1/w)

x³ - 3 x =
(w+1/w)³ - 3(w+1/w) =
w³ + 3 w² (1/w) + 3 w (1/w)² + (1/w)³ - 3 (w+1/w) =
w³ + 3 w + 3 (1/w) + (1/w³) - 3 (w+1/w) =
w³ + (1/w³).

Replacing w³ with z, we get:

z + 1/z = 0

z² = -1

There's no escaping those complex numbers this time!

z = i, -i.

Going with the i solution, we now need to solve

w³ = i

Well, one of the solutions is w = -i. This gives us x = w+1/w = -i + 1/-i = -i + i = 0.

This tells us that (x-0) is a factor of x³-3x, and the rest is routine: x³-3x = x(x²-3) giving solutions x = 0, √3, -√3.

Final Example

Now consider x³ - 3x + 1 = 0.

No Simple Solution

As far as I know, there is no simple way to solve this.

Cardano To The Rescue

Replace x with (w+1/w)

x³ - 3 x + 1 =
(w+1/w)³ - 3(w+1/w) + 1 =
w³ + 3 w² (1/w) + 3 w (1/w)² + (1/w)³ - 3 (w+1/w) + 1 =
w³ + 3 w + 3 (1/w) + (1/w³) - 3 (w+1/w) + 1=
w³ + (1/w³) + 1.

Replacing w³ with z, we get:

z + 1/z + 1 = 0

z² + 1 + z = 0

z = -1/2 + i √3/2, -1/2 - i √3/2

Going with the 1/2 + i √3/2 solution, we now need to solve

w³ = -1/2 + i √3/2

ie w³ = cos(120°) + i sin(120°)

This means the solutions are: w = cos(40°) + i sin(40°), cos(160°) + i sin(160°), cos(280°) + i sin(280°)

These give us x = w+1/w = 2cos(40°), 2 cos(160°), 2cos(280°)

The Quadratic Equation

The general quadratic equation is:

a x² + b x + c = 0.

The idea is to reduce it to another quadratic

y² = T.

We know how to solve this: y= √T, −√T.

Note: if a,b,c are real in the general equation, then T will be real in the reduced equation. I'm emphasizing this because the corresponding statement for the cubic equation is NOT true.

The usual steps are:

Divide by a to get an equation of the form

x² + B x + C = 0 (here B = b/a, C = c/a).

Now replace x with y+k, where k is chosen so that the equation is of the form

y² + D = 0.

This is done by choosing k to satisfy 2k = -B, ie k = -B/2.

Taking T = -D, we now have y² = T.

We're done!

The Cubic Equation

The general cubic equation is:

a x³ + b x² + c x + d = 0

The idea is to reduce it to another cubic

w³ = T.

We know how to solve this.

Note: even if a,b,c,d are real in the general equation, that does NOT mean that T will be real. It could be any complex value.

There isn't that much more to it.

All the important steps have been covered in the examples. The other ideas are present in the quadratic case.

Here's what you do:

Note that the first two steps have direct analogues in the quadratic case.

Firstly divide by a to get an equation of the form

x³ + B x² + C x + D = 0.

Now replace x with y+k, where k is chosen so that the equation is of the form

y³ + p y + q = 0.

This is done by choosing k to satisfy 3k = −B, ie k = −B ⁄ 3.

Now replace y with w+m/w, where m is chosen so that the w and 1/w terms disappear. This is done by choosing m to satisfy p = -3m, ie m = - p/3. This is why I chose p = -3 in the above examples.

We now have a sixth order eqation in w of the form

w³ + R(1/w³) + S = 0.

(or, if you prefer w⁶ + S w³ + R = 0)

Taking z = w³, we have a quadratic with 2 solutions, both of which can be complex. Pick one. Call it T.

The general cubic has been reduced to the cubic

w³ = T

Solve this and reverse the above substitutions.

Nomenclature

The substitution x = w+m/w is sometimes called Vieta's substitution.

A sixth order equation of the form

w⁶ + S w³ + R = 0

is sometimes called a tri-quadratic equation

A third order expression of the form

y³ + p y + q

is sometimes called a depressed cubic

Feedback

If you have corrections, additions, modifications, etc please let me know mailto:walterv@gbbservices.com