Algebra Precalculus – Simplifying Ramanujan-type Nested Radicals

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Ramanujan found many awe-inspiring nested radicals, such as…

$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\sqrt[3]{\sqrt[5]{\frac {32}{5}}-\sqrt[5]{\frac {27}{5}}}=\frac {1+\sqrt[5]{3}+\sqrt[5]{9}}{\sqrt[5]{25}}\tag{3}$$$$\sqrt[3]{(\sqrt{2}+\sqrt{3})(5-\sqrt{6})+3(2\sqrt{3}+3\sqrt{2})}=\sqrt{10-\frac {13-5\sqrt{6}}{5+\sqrt{6}}}\tag{4}$$$$\sqrt[6]{4\sqrt[3]{\frac {2}{3}}-5\sqrt[3]{\frac {1}{3}}}=\sqrt[3]{\sqrt[3]{2}-1}=\frac {1-\sqrt[3]{2}+\sqrt[3]{4}}{\sqrt[3]{9}}\tag{5}$$

And there's more!

Question: Is there a nice algebraic way to denest each radical such as above?

For me, I've only been able to prove such identities by raising both sides to the appropriate exponents and use Algebra to simplify them. But sometimes, that can be very difficult for identities such as $(1)$.

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