<s>[INST] How to denest $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ from scratch? I have seen several questions asking for the proof of
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
However, I want to simplify $\sqrt[3]{\sqrt[3]{2}-1}$ into the form $\sqrt[3]{a}-\sqrt[3]{b}+\sqrt[3]{c}$ and find $a, b, $ and $c$ from scratch.
How would I go about doing it?
Thanks.
 [/INST] For nested cubic-roots of the form $\sqrt[3]{p \sqrt[3]{q}+r}$, there is a systematic denesting method due to Ramanujan.
If the cubic polynomial
$x^3 +a x^2 +bx +c=(x-x_1)(x-x_2)(x-x_3)$ satisfies
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
per Ramanujan
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
To denest $\sqrt[3]{\sqrt[3]{2}-1}$, match it to the right-hand-side of (2)
$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=2$$
which, together with (1), determines $a=-\frac13$, $b=-\frac2{27}$ and $c= \frac8{729}$, or
$$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}= (x-\frac19)(x+\frac29)(x-\frac49)$$
Then, the Ramanujan’s formula (2) yields
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
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