Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38}  - \sqrt[3]{17\sqrt{5}-38}$$

Here's the thing about this question: It is a trick.For most numbers of the form $a+b\sqrt 5$ the cube root is a mess. But some such numbers have nice cube roots and when they do we can find them.
The one thing you have to know is that if $x$ is any number of the form $a+b\sqrt5$ then $x^2, x^3$, etc. also have that form.  For example: $$\begin{align}
(a+b\sqrt 5)^2 & = a^2 + 2ab\sqrt 5 + 5b^2 \\
& = (a^2 + 5b^2) + 2ab\sqrt 5
\end{align}$$
Or similarly $$(a+b\sqrt 5)^3 = (a^3 + 15ab^2) + (3a^2b+5b^3)\sqrt 5.\tag{$\star$}$$
If $17\sqrt5 + 38$ has a simple cube root, it will be something of the form $a+b\sqrt 5$.  From $(\star)$ we would want $$\begin{align}
a^3+ 15ab^2 & = 38 \\
3a^2b+5b^3 &= 17 \\
\end{align}$$
Looking at the first equation we see that maybe $15ab^2 = 30$ is almost 38, and then $a=2, b=1$ would work.  This also works in the second equation. (Or we could look at the second equation first and see right away that the only way to get $3x+5y=17$ is to have $x=4, y=1$, so $a=2, b=1$.) This is called solving equations "by inspection". It is not quite a lucky guess; it is a thoughtful lucky guess based on the hope that the solution will be simple.  But it also wouldn't be hard to try every possibility for $a$ and $b$: they can't be too big because then $a^3+15ab^2$ would be much bigger than 38.
Now we have $a=2, b=1$ we know that $\sqrt[3]{17\sqrt 5+ 38} = 2+\sqrt 5$. We can use the same method to find $\sqrt[3]{17\sqrt 5- 38}$ exactly, and from there the solution is easy.
