Simplifying $\sqrt[5]{1+g+g^3}=\frac {\sqrt{1+g^2}}{\sqrt[10]{5}}$ and similar ones I saw that Ramanujan simplified many radicals such as: 
For $g^5=2$
$$\sqrt[5]{1+g+g^3}=\frac {\sqrt{1+g^2}}{\sqrt[10]{5}}\tag{1}$$
For $g^4=5$
$$\frac {\sqrt[5]{3+2g}-\sqrt[5]{4-4g}}{\sqrt[5]{3+2g}+\sqrt[5]{4-4g}}=2+g+g^2+g^3\tag{2}$$
For $g^5=2$
$$\frac {\sqrt{g+3}+\sqrt{5g-5}}{\sqrt{g+3}-\sqrt{5g-5}}=g^2+g\tag{3}$$
For $g^5=3$
$$\sqrt[3]{2-g^3}=\frac {1+g-g^2}{\sqrt[3]{5}}\tag{4}$$
For $g^5=3$
$$\frac {\sqrt{g^2+1}+\sqrt{5g-5}}{\sqrt{g^2+1}-\sqrt{5g-5}}=g^3+g^2+g+\frac {1}{g}\tag{5}$$
And the list goes on and on... My question is: How do we prove such equations and is there a general method that can be used?
After giving it some thought, I wondered if you can use componendo et dividendo to prove it. Where given the fraction $\frac {a}{b}=\frac {c}{d}$, we have $\frac {a+b}{a-b}=\frac {c+d}{c-d}$. But I don't think we can use that method to find other examples of $(1)$ or $(4)$.

Each equation can be manipulated until it is an equality of polynomials in $g$. For example, putting (1) to the power of $10$, it is equivalent to
$$
  5(1+g+g^3)^2=(1+g^2)^5.
$$
(note that the operation can be reversed because both sides of the original equation are positive).
A polynomial in $g$ can be simplified using $g^{k+5}=2g^k$, until only exponents less than $5$ remain. At this point the expressions should be identical (if they are not, and the coefficients are rational, then equality does not hold). In this case the LHS is
$$\begin{eqnarray*}
  5(1+g+g^3)^2
    &=&5(1+g^2+g^6+2g+2g^3+2g^4)\\
    &=&5(1+4g+g^2+2g^3+2g^4)
\end{eqnarray*}$$
and the RHS is
$$\begin{eqnarray*}
  (1+g^2)^5
    &=&1+5g^2+10g^4+10g^6+5g^8+g^{10}\\
    &=&1+5g^2+10g^4+20g+10g^3+4\\
    &=&5(1+4g+g^2+2g^3+2g^4)
\end{eqnarray*}$$
as required.
