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

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.
