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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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Probably. The best route would be to find out how Ramanujan did that one and then generalize. – anon Sep 4 '11 at 18:10
it was proposed in the JIMS (indian journal) . but i didn't think that ramanujan had give a proof for it . – Prigmin Smith Sep 4 '11 at 18:44
@anon, if any of us could find out how Ramanujan created his mathematics, we wouldn't be spending our time at – Gerry Myerson Nov 25 '11 at 21:58
@PrigminSmith, homework? Really? – Gerry Myerson Nov 25 '11 at 22:00
My (superficial) understanding is that Ramanujan had a much greater willingness than we do to pursue formal manipulations like this for months or years at a time without any immediate payoff in sight; eventually, he developed a superhuman talent for it. I think you'd have a hard time publishing something like the equation you give today without also suggesting a pretty direct application, which may be short-sighted on the part of modern mathematics -- if Ramanujan hadn't studied this sort of thing, the eventual applications to analytic number theory never would have emerged. – Daniel McLaury Nov 25 '11 at 23:40

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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Note: I think it should be possible to choose the right hand side arbitrarily, square it and get a suitable expression for the left hand side. This way, one should be able to get as many such identities as we like. One clever trick in doing so, however, is to choose the coefficients of $u^2$, $uv$ and $v^2$ carefully, so that the $u^2 v^2$ term vanishes on squaring. – Srivatsan Nov 26 '11 at 0:34
It's better to write the identity as: $3\sqrt{m\sqrt[3]{4(m-2n)}+n\sqrt[3]{4m+n}}=\left| \sqrt[3]{(4m+n)^{2}}+\sqrt[3]{4(m-2n)(4m+n)}-\sqrt[3]{2(m-2n)^{2}} \right|$ – Ng Chung Tak Jan 16 at 19:37

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