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.