Ramanujan found that
$$\begin{align*} & \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}\\ =&\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9}-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9}+\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9} \end{align*}$$ for arbitrary $m$ and $n$. The problem is that I am not sure how to prove it.
Question: Is there a way to prove it?
Similarly to the Ramanujan's formula for $$\sqrt{\sqrt[3]{a}+\sqrt[3]{b}}$$ I tried starting with a polynomial and deriving above formula through some clever manipulation. Unfortunately, none of that worked. If I were to do it that way, the terms in the polynomial would have to be a square root.
The book provides its way of proof by cubing both sides brute-force and slugging out the difficult algebra. Is there another way to prove it?
Denote $$c_1=\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9},\>\>\> c_2=-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9},\>\>\> c_3=\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9} $$
and it is straightforward to verify that
\begin{align} c_1c_2c_3 &=- \frac{1}{9}(m-n)(2m+n)(m+2n)\\ c_1^3+c_2^3+c_3^3 &= \frac{1}{3}(m^3+6m^2n+3mn^2-n^3)\\ c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 &= \frac{1}{3} c_1c_2c_3(m^3-3m^2n-6mn^2-n^3)\\ \end{align} Next, let $$A=c_1+c_2+c_3, \>\>\> B=c_1c_2+c_2c_3+c_3c_1,\>\>\>C= c_1c_2c_3$$ and evaluate \begin{align} A^3 &= 3AB+ c_1^3+c_2^3+c_3^3-3c_1c_2c_3\\ &=3AB+ (m^3+3m^2n-n^3)\tag1 \\ B^3 &= 3c_1c_2c_3 AB + c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3-3(c_1c_2c_3)^2\\ &=3(AB)C +(m^3-3mn^2-n^3)C\tag2 \end{align} and their product $$ A^3B^3 = 9(AB)^2C-27(AB)C^2+27C^3 -\frac13C(m^2+mn+n^2)^3\tag3 $$ where the followings are recognized in arriving at (3) \begin{align} & (m^3+3m^2n-n^3)+(m^3-3mn^2-n^3)=-9C \\ & (m^3+3m^2n-n^3)(m^3-3mn^2-n^3)=27C^3-\frac13(m^2+mn+n^2)^3\\ \end{align} Rearrange (3) $$(AB-3C)^3=-\frac13C(m^2+mn+n^2)^3$$ and substitute $AB$ via (1) to obtain the equation for $A$ $$[A^3-(3mn^2+n^3-m^3)]^3 = -9C(m^2+mn+n^2)^3$$ which leads to the Ramanujan formula $$A= \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3} $$