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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 $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}-\sqrt[3]{b}+\sqrt[3]{c}$$ and find $a, b, $ and $c$ from scratch.

How would I go about doing it? Thanks.

Quanto
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Mike Smith
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    I worked on this problem for over an hour, and I got nowhere. This is proving a tough nut to crack. I suggest that if you get no response that you spend an hour or so working on it, and then edit your query to include all of your work, regardless of how close you think that you are to a solution. You will get a much better response from mathSE reviewers more talented than I am if they see that you have made a serious effort and documented this effort in your query. – user2661923 Oct 25 '20 at 04:23
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    You won't succeed as there are plenty possibilities for $a$, $b$, and $c$. – Michael Hoppe Oct 25 '20 at 14:52
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    I believe this one is due to Ramanujan (who else?), and you can see in https://faculty.math.illinois.edu/~berndt/articles/radicals.pdf top of page 5 a generalization that derives it, as well as some similar identities. Anyway, https://math.stackexchange.com/questions/871639/denesting-radicals-like-sqrt3-sqrt32-1 covers your last question. The obvious answer is to cube both sides, leading to a system of equations in $a,b,c$. Without constraints this may not even be solvable, and when it is, may be really difficult to solve. – LPenguin Oct 25 '20 at 14:52
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    Nested radicals like this are indeed hard, if not impossible to solve by hand. So It is best to use a general identity that deals with the de-nesting of form $\sqrt[^3]{A+\sqrt[^3]{B}}$ (which is present in a link by @LPenguin). And also this is what Ramanujan intended in his question: https://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/question/q682.htm – Mourad Oct 25 '20 at 15:53

1 Answers1

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Below is an established procedure to denest the cubic-roots of the form $\sqrt[3]{p \sqrt[3]{q}+r}$. Given the cubic equation $$x^3 +a x^2 +bx +c=(x-x_1)(x-x_2)(x-x_3)$$

the Ramanujan identity $$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag1$$

holds, if the coefficients satisfy $$b +a c^{1/3} +3c^{2/3}=0\tag2$$

Thus, to denest $\sqrt[3]{\sqrt[3]{2}-1}$, match it to the right-hand-side of (1)

$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=2$$

which, together with (2), 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 formula (1) yields

$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$

Quanto
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