This starts my newest blog, one devoted to mathematics and science, but mostly to mathematics. I called it Hilbert's Hotel after the fabulous hotel that was constructed with an infinite amount of space that will hold a countable infinity of guests, and there is always room for one more, since if John Q. Onemoreguest should arrive, we just tell everybody to move out of their rooms and into the room with the next highest number; e.g., from 107 to 108. Then we can fit John in room 1.

Today I got a question to the math help website I belong to, Mathnerds, concerning cubic equations. These have fascinated me since I was small. There was a formula for quadratic equations, but how about cubics? The best way to understand these is to first of all convert the general cubic

ax³ + bx² + cx + d = 0

into one of the form
x³ + px + q = 0

by dividing everything by a and then by applying the transform y = x - a/3 (and then of course x = y). Let w be an imaginary cube root of 1; i.e., w = -1/2 + sqrt(-3)/2. Compute this product:

(x - u - v)(x - wu - w²v)(x - w²u - wv)

You get:

x³ - 3uvx - (u³+v³)

which is exactly in the form of the reduced cubic equation. Then set

-3uv = p

p³ + q³ = -q

and solve by solving for v in the first equation, and plugging into the second, yielding a quadratic in p³. Solve this using the formula, and select a cube root of the result for u. Then solve for v in the first equation, and then the roots are

u + v
wu + w²v
w³u + wv

This is essentially the method of Lagrange resolvents, and the question in Mathnerds asked about those. Another interesting way of looking at it is through Dan Kalman's circulants.

There is also a discriminant D. If D is 0, the roots are probably rational and two of them are the same. If D > 0, then there is one real root and two imaginary ones, and one can actually come up with a formula for the real root, but it will be somewhat complex looking, being a sum of the cube roots of two quadratic expressions.

The really confusing case is the irreducible one, where D < 0, and my favorite equation here is x3 - 3x + 1 = 0. If you use the formula above, you get w^1/3 + w^2/3 for one of the roots. The thing is, this is real, yet it is expressed in imaginary quantities. I searched all over when I was young for a real solution, until I realized that there are none. There is no way of expressing the roots of this equation in real radicals. If you are willing to use trig functions, there is a way: 2cos(40 degrees) or 2 cos (2*π/9) is one of the roots, for instance.

Cubic equations were originally solved by the Italian mathematicians Tartaglia and Cardano, and Galois added a lot to the theory by associating with equations and field extensions a structure called a Galois group, which for most cubic equations is the group of permutations of three objects, known as S₃. Most of the time when confronted with one, one uses Newton's method to approximate the real roots. But cubic equations solved algebraically have a fascination with me and they will continue to do so.