Simplifying Ramanujan-type Nested Radicals 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 $(1)$.

Considering
$$(\sqrt[4]{a} \pm \sqrt[4]{b})^{4}=
a+b+6\sqrt{ab} \pm 4\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})$$
Making a factor in the form of $\sqrt{m}+\sqrt{n}$,
$$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{a}}{\sqrt{b}}
\implies \frac{a}{b}=5$$
Then
$$\left( \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1} \right)^{4}=
\frac{(\sqrt{5}+1)(6+4\sqrt[4]{5})}
     {(\sqrt{5}+1)(6-4\sqrt[4]{5})}=
\frac{3+2\sqrt[4]{5}}
     {3-2\sqrt[4]{5}}$$
Note on the symmetric roles of $a$ and $b$:
$$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{b}}{\sqrt{a}}
\implies \frac{b}{a}=5$$
which gives the same result.

Hence, there's no other cases similar to $(2)$.

