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Question Number 59626 by naka3546 last updated on 12/May/19

$$Rationalizethedenominatorof$$$$\frac{2}{1-\sqrt{2+\sqrt[3]{4}}}$$

Answered by MJS last updated on 12/May/19

$$\mathrm{first}\mathrm{use}\mathrm{this}\mathrm{formula}$$$$\frac{1}{a-\sqrt{b}}=\frac{a+\sqrt{b}}{(a-\sqrt{b})(a+\sqrt{b})}=\frac{a+\sqrt{b}}{a^2-b}$$$$\mathrm{then}\mathrm{use}\mathrm{this}\mathrm{formula}$$$$\frac{1}{a+\sqrt[3]{b}}=\frac{(a+\sqrt[3]{b}(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i}))(\mathrm{a}+\sqrt[3]{b}(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}))}{(a+\sqrt[3]{b})(a+\sqrt[3]{b}(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i}))(\mathrm{a}+\sqrt[3]{b}(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}))}=$$$$=\frac{a^2-a\sqrt[3]{b}+\sqrt[3]{b^2}}{a^3+b}$$

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