Simplifying $x-\dfrac1x$, where $x = \sqrt[3]{54+38\sqrt{2}+3\sqrt{645+456\sqrt{2}}}$

Asked Oct 19 '22 at 08:28

Active Oct 19 '22 at 08:48

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This question arises from this question: How do I solve for $x-\frac{1}{x}$ here?

One of the answers recommends I use the quadratic formula to solve for $u$, and cube root that to get $x$. I'm getting $$x = \sqrt[3]{54+38\sqrt{2}+3\sqrt{645+456\sqrt{2}}}$$ I don't really need the value of x: but I do need the value of $x-\dfrac{1}{x}$. So really, the radical I need to simplify is this:

$$\sqrt[3]{54+38\sqrt{2}+3\sqrt{645+456\sqrt{2}}} - \frac{1}{\sqrt[3]{54+38\sqrt{2}+3\sqrt{645+456\sqrt{2}}}}$$

edited Oct 19 '22 at 08:48

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asked Oct 19 '22 at 08:28

avighnac

This is just $3+2 \sqrt{2}$ – Claude Leibovici Oct 19 '22 at 08:46
@ClaudeLeibovici yeah, it is! Could you post an answer w/ your working and thought process? – avighnac Oct 19 '22 at 08:49
8

Just mental calculation ! (joke) – Claude Leibovici Oct 19 '22 at 08:51
My first try, which might not succeed, would be $$\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2.$$ My second try, which also might not succeed, would be $$\left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right).$$ – user2661923 Oct 19 '22 at 09:05
@user2661923 you mean substitute $x$ in the right part for both cases? – avighnac Oct 19 '22 at 09:10
@avighnac No, I mean that you apparently have a value for both $~x~$ and $~\displaystyle \frac{1}{x}.~$ I suggest using both values. – user2661923 Oct 19 '22 at 09:18
@user2661923 ah, okay. – avighnac Oct 19 '22 at 09:22

Simplifying $x-\dfrac1x$, where $x = \sqrt[3]{54+38\sqrt{2}+3\sqrt{645+456\sqrt{2}}}$

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