<s>[INST] Simplifying $\sqrt[3]{a\pm\sqrt{b}}$ Let
$$x=\sqrt{a\pm\sqrt{b}}$$
We know that
$$x=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
But, what about cubic root?
Let
$$y=\sqrt[3]{a\pm\sqrt{b}}$$
Is there any formula to find $c$ and $d$ such that $c,d\in\mathbb{Q}$ and $c\pm\sqrt{d}=y$ if $c$ and $d$ exists?
For example, let
$$a=\sqrt[3]{45+\sqrt{1682}}$$
It can be solved factoring terms:
$$a=\sqrt[3]{45+29\sqrt{2}}=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}=3+\sqrt{2}$$
Is there any formula for cubic root like square root?
 [/INST] There isn't a formula for $\sqrt[3]{a\pm\sqrt{b}}$ The best method that I know is simplifying $\sqrt{b}$ (if possible) and assuming that it can be denested into $x+y\sqrt{b}$.
More generally, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}$$
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