Show how to find the cubic root

Question

Show how to find the cubic root of surds of the form $A+\sqrt[3]{B}$, and deduce that

$$\sqrt[3]{{\sqrt[3]{2} - 1}}= \sqrt[3]{1 \over 9}-\sqrt[3]{2 \over 9}+\sqrt[3]{4 \over 9}$$

Answer 1

If you are talking about Ramanujan's radicals, that's a different story, and one that has been very nicely addressed. By Johannes Blomer.

Answer 2

Well, it is must be useful to think in something.

${\sqrt[3]{{\sqrt[3]{2} - 1}}}.\frac{\sqrt[3]{{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}}}{\sqrt[3]{{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}}}=\frac{1}{\sqrt[3]{{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}}}$

$=\frac{1}{\sqrt[3]{{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}}}.\frac{\sqrt[3]{3}}{\sqrt[3]{3}}=\frac{\sqrt[3]{3}}{\sqrt[3]{{3.(\sqrt[3]{2^2})+3.(\sqrt[3]{2})+3}}}$

$\frac{\sqrt[3]{3}}{\sqrt[3]{{(\sqrt[3]{2}+1)^3}}}=\frac{\sqrt[3]{3}}{(\sqrt[3]{2}+1)}.\frac{{{{(\sqrt[3]{4}-\sqrt[3]{2}+1)}}}}{{{{(\sqrt[3]{4}-\sqrt[3]{2}+1)}}}}=\frac{\sqrt[3]{3}.(\sqrt[3]{4}-\sqrt[3]{2}+1)}{2+1}=\sqrt[3]{\frac{4}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{1}{9}}$