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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?

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Looks like you divided the right side by $\sqrt[3]3$. The remaining $\sqrt[3]2$ should come from the numerator.

$$2+\sqrt[3]2-\sqrt[3]4=\sqrt[3]2(\sqrt[3]4+1-\sqrt[3]2)$$

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  • $\begingroup$ Ah! I see what I did wrong! Dumb mistake on my part! :/ $\endgroup$ – Frank Jun 2 '16 at 4:22

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