Any other cases similar to $\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac 12 (3+\sqrt[4]{5}+\sqrt{5}+\sqrt[4]{125})$ Ramanujan gave many curious identities, one of which was $$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}=\frac 12(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125})\tag1$$
And another one:$$\sqrt[3]{\sqrt[5]{\frac {32}5}-\sqrt[5]{\frac {27}5}}=\sqrt[5]{\frac 1{25}}+\sqrt[5]{\frac {3}{25}}-\sqrt[5]{\frac {9}{25}}\tag2$$


Question: Is there some algebraic method to prove $(1)$ and $(2)$, and can that method be generalized? And more specifically, can both $(1)$ and $(2)$ be represented as a sum of radicals of the same degree? Such as $\sqrt[4]{A+B\sqrt[4]{C}}=\sqrt[4]{X}+\sqrt[4]{Y}+\sqrt[4]{Z}$.


Also, I do realize that for $(1)$, you can raise everything to the fourth power and simplify to get an identity (such as $1=1$), but it doesn't provide insight on why or how the equation holds.

The first identity of (1) can not be generalized, namely $5$ is the only integer satisfying (1). To see this, let
$$ \sqrt[4]{\frac{a+b\sqrt[4]{p}}{a-b\sqrt[4]{p}}}=\frac{\sqrt[4]{p}+q}{\sqrt[4]{p}-q} \tag{3}$$
where $p,q$ are integers such that $\sqrt[4]{p},\sqrt{p}$ are not integers and $a,b,$ are rationals. Then (3) is equivalent to
$$ (a+b\sqrt[4]{p})(\sqrt[4]{p}-q)^4-(a-b\sqrt[4]{p})(\sqrt[4]{p}+q)^4=0 $$
or
$$ \sqrt[4]{p}\bigg\{2q(-2a+3qb)\sqrt{p}+\big[-4q^3a+(p+q^4)b\big]\bigg\}=0. $$
Let
$$ \bigg\{\begin{array}{ll}
-2a+3qb&=&0,\\
-4q^3a+(p+q^4)b&=&0,
\end{array}$$
which is a system of linear homogeneous equations. It has nonzero solutions iff
$$ \det\bigg(\begin{matrix}
-2&3q\\
-4q^3&p+q^4
\end{matrix}\bigg)=0 $$
from which one obtains $p=5q^4$ and $a=\frac32qb$. Thus (3) becomes
$$ \sqrt[4]{\frac{3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1}. $$
