Dear Uncle Colin,

Apparently $\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}$ is an integer. Really?

Can A Root Define A Number? Oo!

Hi, CARDANO, and thanks for your message!

Let’s eyeball it first: $2 + \sqrt{5}$ is about 4.2, which has a cube root somewhere about 1.6, I’d have thought; $2-\sqrt{5}$ is about -0.2, and its cube root would be about -0.6, so if it’s an integer, it’s probably 1. That’s all well and good if you’re a pirate, but what if you want a proper answer?

I tend to approach these by cubing.

Let $a = \sqrt[3]{2+\sqrt{5}}$ and $b = \sqrt[3]{2 - \sqrt{5}}$.

Then $(a+b)^3 = a^3 + 3ab(a+b) + b^3$, if you combine the middle terms.

We know what $a^3 + b^3$ is – it’s $(2+\sqrt{5}) + (2-\sqrt{5})$, which is 4, which makes things easier.

What about $ab$? Well, $a^3b^3$ is a difference of two squares, giving $a^3b^3 = -1$, so $a^3b^3 = -1$.

That means $(a+b)^3 = 4 - 3(a+b)$.

If we let $x = (a+b)$, that’s a cubic: $x^3 + 3x - 4 = 0$

How do you solve a cubic? Here, it’s easy enough to spot that $x=1$ is a solution. Are there others? Let’s factorise to get $(x-1)(x^2 +4x + 4)$, so $x=-2$ (twice) is the other solution. Looking at the relative sizes of $a$ and $b$, our solution is positive, so it must be 1.

Hope that helps!

- Uncle Colin

* Thanks to @aw_mair for pointing out a typo. Edited 2022-10-31 to fix it.