Prove Ramanujan's formula for nested cubic roots $\sqrt[3]{{a}+b\sqrt[3]{r}}$ 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?
 A: 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
$$A= \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}
$$
