Fresh from finding the continued fraction of $\sqrt {19}$, I wondered “how hard can it be to find something like $\sqrt [3]{2}$?”.

Much harder, it turns out. But! There’s some interesting maths along the way.

Following a similar scheme to last time, I noticed that $r = \sqrt [3]{2}$ is a bit more than 1, so it can be written as $r = 1 + \frac {1}{x_1}$.

Rearranging that gives $x_1 = \frac {1}{r - 1}$, which is where the trouble starts: the difference of cubes isn’t so nice as the difference of squares. It’s not so bad here, though: it’s clear enough that $r^3 - 1 = (r-1)(r^2 + r + 1)$.

That means we can write $\frac {1}{r-1}$ as $r^2 + r + 1$, noting that $r^3-1 = 1$.

Let’s take it as read that the Mathematical Ninja has told us that $\sqrt [3]{2} \approx \frac {5}{4}$ and that $\sqrt [3]{2}^2 \approx \frac {8}{5}$ via some sort of binomial trickery. That tells us that our $x_1$ is between 3 and 4, which means our continued fraction starts $\left [1;3\right ]$ and we’re left with $x_2 = \frac {1}{r^2 + r - 2}$.

At first glance, that’s a bit brutal: however, the denominator factorises as $(r+2)(r-1)$, so we can infer that the multiplier is $(r^2 - 2r + 4)(r^2 + r + 1)$. Remembering that $r^3=2$ turns this into $3r^2 + 4r+2$, and the denominator into 10.

How much is that? The top is a little over 11, so our next value is 1 and our next denominator is $3r^2 + 4r - 8$ – but this doesn’t factorise.

Instead, we need to work through some tedious algebra to find $(ar^2 + br + c)$ such that $(ar^2 + br + c)(3r^2 +4r -8)$ is an integer. This involves multiplying the whole thing out, replacing all of the $r^3$ terms with 2, and making sure the $r^2$ and $r$ terms vanish. I get $a=c=4$ and $b=5$, which gives a denominator of 30.

I’m not going to go any further with it, but I think this is something to do with field extensions – it’s neat that it always seems to be possible to rationalise the cubic denominator, even if it’s not pleasant. I wonder if there’s an easier process?