simplify $\sqrt[3]{11+\sqrt{57}}$ I read in a book (A Synopsis of Elementary Results in Pure and Applied Mathematics) that the condition to simplify the expression $\sqrt[3]{a+\sqrt{b}}$ is that $a^2-b$ must be a perfect cube.
For example $\sqrt[3]{10+6\sqrt{3}}$ where $a^2-b
=(10)^2-(6 \sqrt{3})^2=100-108=-8$ and $\sqrt[3]{-8} = -2$
So the condition is satisfied and $\sqrt[3]{\sqrt{3}+1}^3=\sqrt{3}+1$.
But the example $\sqrt[3]{11+\sqrt{57}}$ where $a^2-b =
(11)^2-57=121-57=64$ and $\sqrt[3]{64}=4$ so the condition is satisfied.
But I can’t simplify this expression. Please help us to solve this problem. Note: this situation we face it in many examples
 A: I imagine you want to write, with integer $u$ and $v$,
$$11+\sqrt{57}=(u+\sqrt{v})^3$$
So that the cube root would simplify.
Hence
$$u^3+3u^2\sqrt{v}+3uv+v\sqrt{v}=11+\sqrt{57}$$
So you would need
$$\begin{eqnarray}
u(u^2+3v)&=&11\\
v(3u^2+v)^2&=&57
\end{eqnarray}$$
But $57=3 \cdot 19$ has no square factor, so $3u^2+v=\pm1$, and $v=57$. But then the first equation can't hold, since $u^2+3v\geq 3\cdot57=171$.
So it's in fact not possible to simplify this way.
A: Here’s the story: The real quadratic number field $\mathbb Q(\sqrt{57}\,)$ fortunately has class number $1$ (I looked it up on the internet), so that it is relatively easy to do arithmetic there. The fundamental unit is $\varepsilon=151+20\sqrt{57}$, which has norm $+1$, so that $1/\varepsilon$ is $151-20\sqrt{57}$. The other thing we need to do is know how the prime $2$ behaves there. Since $57\equiv1\pmod8$, the prime $2$ splits, indeed $-2=zz'$, where $z=(7+\sqrt{57}\,)/2$, and $z'$ is the same thing but with a minus sign.
Now here’s how to factor $11+\sqrt{57}$:
$$
11+\sqrt{57}=2\frac{11+\sqrt{57}}2=-zz'\frac{11+\sqrt{57}}2\,,
$$
but, setting $(11+\sqrt{57}\,)/2=w$, we compute that $w/z^4=151-20\sqrt{57}=1/\varepsilon$. Thus the factorization of our original number is
$$
11+\sqrt{57}=-\varepsilon^{-1}z^5z'\,,
$$
which is definitely not a cube in our field. In case you look instead at $11-\sqrt{57}$, it will get expressed as $-\varepsilon z{z'}^5$, still not a cube.
A: It is not a sufficient condition (I don't know if it's necessary). Not all expressions of the form $\sqrt[3]{a+\sqrt{b}}$, satisfying the condition that $a^2-b$ is a perfect cube, can be simplified.
A: I think we want a "-" sign not a "+" sign.
Squaring under the radical gives 64. Then the square root and a cube root gives 2.
