Ramanujan's identity involving cube roots I have found this pretty identity involving cube roots, first stated by Ramanujan: 
$$
\sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- (2(m-2n))^{\frac23}.
$$
So I ask if there is any method to create similar identities? 

The identity seems incorrect; instead it should read $(\ast)$
$$
\color{Red}{3} \cdot \sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- \color{Red}{(2(m-2n)^2)^{\frac13}}.
$$
I might be asking for trouble by being the heretic =), but I find this particular identity quite unremarkable. 
The key is the substitution 
$$
u^3 = 4m+n, \quad v^3 = m-2n,
$$
motivated by the repeated occurrences of $(4m+n)^{1/3}$ and $(m-2n)^{1/3}$. Then
$$
m = \frac{2u^3  + v^3}{9}, \quad n = \frac{u^3 - 4v^3}{9}.
$$
Plugging in these in $(\ast)$,
$$
3 \cdot \sqrt{\frac{4^{1/3}(2u^3v+v^4) + (u^4 - 4uv^3)}{9}} \mathop{\stackrel{\color{Blue}{(?)}}{=}}
u^2 + 4^{1/3} uv - 4^{1/3} v^2
$$
$$
\iff u^4 + 2 \cdot 4^{1/3} \cdot u^3v - 4uv^3 + 4^{1/3} v^4 \mathop{\stackrel{\color{Blue}{(?)}}{=}} 
\Big(u^2 + 4^{1/3} uv - 4^{1/3} v^2 \Big)^2,
$$
which can be verified by squaring the right hand side. 
