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 think we want a "-" sign not a "+" sign.
Squaring under the radical gives 64. Then the square root and a cube root gives 2.
