I was messing around with nested radicals, and I came up with $\sqrt[3]{6\sqrt[3]{2}-6}=\frac {2+\sqrt[3]{2}-\sqrt[3]{4}}{\sqrt[3]{3}}$.
The $\sqrt[3]{2}$ and $\sqrt[3]{4}$ seemed familiar to another nested radical exmaple: $\sqrt[3]{\sqrt[3]{2}-1}=\frac {1-\sqrt[3]{2}+\sqrt[3]{4}}{\sqrt[3]{9}}$.
So I decided to give it a try. I divided both sides of $\sqrt[3]{6\sqrt[3]{2}-6}$ by $\sqrt[3]{6}$ and got that $\sqrt[3]{\sqrt[3]{2}-1}=\frac {2+\sqrt[3]{2}-\sqrt[3]{4}}{\sqrt[3]{9}}$
But this isn't what Ramanujan had... Something went wrong. Any ideas?
Looks like you divided the right side by $\sqrt[3]3$. The remaining $\sqrt[3]2$ should come from the numerator.
$$2+\sqrt[3]2-\sqrt[3]4=\sqrt[3]2(\sqrt[3]4+1-\sqrt[3]2)$$