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Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.

I did calculate inverse elements in field extensions of degree $2$ over $\mathbb{Q}$, but am somehow stuck with this problem, any help is very much appreciated!

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You can carry out the "rationalise the denominator" in two steps.

First multiply $\frac{1}{5+2\sqrt{2}-3\sqrt{3}}$ by $\frac{5+2\sqrt{2}+3\sqrt{3}}{5+2\sqrt{2}+3\sqrt{3}}$ and get $\frac{5+2\sqrt{2}+3\sqrt{3}}{6+20\sqrt{2}}$.

Now multiply by by $\frac{6-20\sqrt{2}}{6-20\sqrt{2}}$ and get $\frac{(5+2\sqrt{2}+3\sqrt{3})(6-20\sqrt{2})}{-764}$.

Multiply out the numerator and you're done.

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  • $\begingroup$ Thank you! This is a great trick i will definitly keep in mind! Final solution that can be obtained by either one of the ways: $\frac{25}{382}+\frac{22}{191}\sqrt{2}-\frac{9}{382}\sqrt{3}+\frac{15}{191}\sqrt{6}$ $\endgroup$ – user526159 May 18 '18 at 10:29
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Hint

Find $a,b,c,d\in\mathbb Q$ s.t. $$\Big(a+b\sqrt 2+c\sqrt 3+d\sqrt 2\sqrt 3\Big)\Big(2\sqrt 2-3\sqrt 3+5\Big)=1.$$

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  • $\begingroup$ How is this supposed to help? You just restated the question. $\endgroup$ – tomasz May 17 '18 at 12:48
  • $\begingroup$ @tomasz: It's not difficult to solve this equation. $\endgroup$ – Surb May 17 '18 at 13:24
  • $\begingroup$ Maybe so, but that is what OP is asking for help with. $\endgroup$ – tomasz May 17 '18 at 16:29
  • $\begingroup$ Thank you! What a simple way. Definitly worked really good! $\endgroup$ – user526159 May 18 '18 at 10:24
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Let $\beta = 2\sqrt{2}-3\sqrt{3}+5$. Then $(\beta-5)^2 = 35-12\sqrt{6}$ and $((\beta-5)^2-35)^2 = 864$, and so $$ \beta^4 - 20 \beta^3 + 80 \beta^2 + 200 \beta = 764 $$ Writing this as $$ \beta(\beta^3 - 20 \beta^2 + 80 \beta + 200) = 764 $$ gives $\beta^{-1} = \frac{1}{764}(\beta^3 - 20 \beta^2 + 80 \beta + 200)$.

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