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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?

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    Incidentally, I believe that they've found a notebook of Gauss's (?) where he apparently spent several months using the cubic and quartic formulas to get complicated expressions that were equivalent to rational numbers, presumably in hopes of identifying an algorithm for simplifying such expressions.2011-11-25

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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.

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    It's better to write the identity as: $3\sqrt{m\sqrt[3]{4(m-2n)}+n\sqrt[3]{4m+n}}=\left| \sqrt[3]{(4m+n)^{2}}+\sqrt[3]{4(m-2n)(4m+n)}-\sqrt[3]{2(m-2n)^{2}} \right|$2016-01-16