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?

  • $\begingroup$ i think it is usually not the case that there exist such $c,d, \in \mathbb{Q} $ , therefor it might be hard to find such a formula $\endgroup$ – supinf Aug 14 '14 at 10:44

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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