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While trying to answer this question, I got stuck showing that

$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$

The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.

That got me thinking... if you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd?

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    If you take 26+15√3 and multiply it by its conjugate, 26−15√3, then you get its **norm**, the perfect cube 1. This lets you know that 26+15√3 could be a perfect cube, and that its cube root would have its norm equal to the cube root of 1. Sure enough 2+√3 times 2−√3 is equal to 1. However, this isn't sufficient, since maybe 2+√3 doesn't have a cube root (or if it does, then maybe its cube root has no cube root, etc.). Finding elements of norm 1 is called **Pell's Equation** and would be one way to check if 2+√3 has another cube root, or if it is finished.2011-01-04

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