Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38}  - \sqrt[3]{17\sqrt{5}-38}$$
 A: Let $u=\sqrt[3]{38+17\sqrt{5}}$, $v= \sqrt[3]{38-17\sqrt{5}}$
From the equation
$$u+v=n$$
and cubing,
we obtain
$$u^3+v^3+3uv(u+v)=n^3$$
that is
$$76-3n=n^3$$
A root of this equation is $4$. In fact, it has no more real roots.
So we know that $u+v=4$, but we need $u-v$. We also know that $uv=-1$. Then
$$u-v=\sqrt{(u+v)^2-4uv}=\sqrt{16+4}=2\sqrt 5$$
A: 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.
A: The expression strangely reminds of Cardano's formula for the cubic equation
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}.$$
Identifying, we have 
$$q=-76,p=3,$$
and $x$ is a root of $$x^3+3x-76=0.$$
By inspection (trying values close to $\sqrt[3]{76}$), $\color{green}4$ is a solution.
As $$x^3+3x-76=(x-4)(x^2+4x+19),$$ the other roots are complex.
A: It is $2\sqrt{5}$. Note that $(17\sqrt{5}+38)^{\frac{1}{3}}=2+\sqrt{5}$. 
A: $$\sqrt[3]{\ 17\sqrt{5}+38}  - \sqrt[3]{17\sqrt{5}-38}=$$
$$\sqrt[3]{8+12\sqrt{5}+30+5\sqrt{5}}  - \sqrt[3]{-8+12\sqrt{5}-30+5\sqrt{5}}=$$
$$\sqrt[3]{8+12\sqrt{5}+6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}  - \sqrt[3]{-8+12\sqrt{5}-6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}=$$
$$\sqrt[3]{\left(\sqrt{5}+2\right)^3}  - \sqrt[3]{\left(\sqrt{5}-2\right)^3}=$$
$$\left({\left(\sqrt{5}+2\right)^3}\right)^{\frac{1}{3}}  - \left({\left(\sqrt{5}-2\right)^3}\right)^{\frac{1}{3}}=$$
$$\left(\sqrt{5}+2\right)  -\left(\sqrt{5}-2\right)=$$
$$\left(\sqrt{5}+2\right)  +\left(2-\sqrt{5}\right)=$$
$$2+2+\sqrt{5}-\sqrt{5}=2+2=4$$
A: Start by noticing that $(2 + \sqrt{5})^{3} = 38 + 17 \sqrt{5}$ and $(2 - \sqrt{5})^{3} = 38 - 17 \sqrt{5}$. Now,
\begin{align}
\sqrt[3]{\ 17\sqrt{5}+38}  - \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} - \sqrt[3]{(\sqrt{5} - 2)^{3}} \\
&= (2 + \sqrt{5}) - (-2 + \sqrt{5}) \\
&= 4.
\end{align}

What seems to have happened is that most answers presented here follow the pattern:
\begin{align}
\sqrt[3]{\ 17\sqrt{5}+38}  + \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} + \sqrt[3]{(\sqrt{5} - 2)^{3}} \\
&= (2 + \sqrt{5}) + (-2 + \sqrt{5}) \\
&= 2 \sqrt{5}.
\end{align}
A: \begin{align} 
x&=\sqrt[3]{17\sqrt{5}+38}  - \sqrt[3]{17\sqrt{5}-38}
\end{align}
Note that 
\begin{align}
17\sqrt{5}-38&=\frac{1}{17\sqrt{5}+38}.
\end{align}
Let $a=\sqrt[3]{17\sqrt{5}+38}$.
Then we have 
\begin{align}
x^3&=\left(a-\frac1a\right)^3
\\
x^3&= a^3-3a+\frac3a-\frac{1}{a^3}
\\
x^3+3\left(a-\frac1a\right)&= a^3-\frac{1}{a^3}
\\
x^3+3x-76&=0
\\
(x-4)(x^2+4x+19)=0,
\end{align}
The only real solution $x=4$ is the answer.
