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I am trying to express the inverse of $a+b \sqrt{2}+c \sqrt{3} + d \sqrt{6}$ (given $a, b, c, d \in \mathbb{Q}$) in the form $e+f \sqrt{2}+g\sqrt{3}+h\sqrt{6}$ (where $e, f, g, h \in \mathbb{Q}$).

I could come up with a long way to do this by solving 4 linear equations to find $e, f, g, h$ which we get by expading the following,

$(a+b \sqrt{2}+c \sqrt{3} + d \sqrt{6})(e+f \sqrt{2}+g\sqrt{3}+h\sqrt{6})=1$ and equating the rational term equal to 1 and the rest of the terms equal to $0$.

But, I am now trying to find a more creative (elegant, generalisable, insightful) way to do this.

For this I tried a few things which did not work (unecessary detail?).

I guessed, $(a+b \sqrt{2}+c\sqrt{3}+d\sqrt{6})(a+b \sqrt{2}+c\sqrt{3}-d\sqrt{6})(a+b \sqrt{2}-c\sqrt{3}+d\sqrt{6})(a-b \sqrt{2}+c\sqrt{3}+d\sqrt{6}) \cdots$

(all the terms with all possible signs), but this did not work as I checked with a special case.

I guessed that since a similar thing works for the rationalising factor of $\sqrt{a}+\sqrt{b}+\sqrt{c}$ which has $(\sqrt{a}+\sqrt{b}-\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})$ (I was motivated to this from heron's formula for area of triangle).

So, can anyone help me? Please. By the way I am in high school so I don't know anything about field splittings and so on.

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Basically, you almost guessed an ok trick:$$\frac1{a+b\sqrt2+c\sqrt3+d\sqrt6}=\\=\frac1{a+b\sqrt2+(c+d\sqrt2)\sqrt3}=\\=\frac{a+b\sqrt2-c\sqrt3-d\sqrt6}{(a+b\sqrt2)^2-3(c+d\sqrt2)^2}=\cdots$$ Can you continue, now that there's only one root in the denominator?

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    $\begingroup$ Ooh! Why did I not think of that. yes I can now finish it off, by writing the new denominator as $m+n\sqrt{2}$ and rationalising with $m-n\sqrt{2}$ $\endgroup$ – Subham Jaiswal Mar 5 '16 at 18:56
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    $\begingroup$ @AmyWinehouse Precisely. I had almost forgotten myself. $\endgroup$ – user228113 Mar 5 '16 at 18:57
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The idea with multiplying the same expression with all possible signs is good, but requires a lot of multiplying, so it is easy to make a mistake. It turns out that it is sufficient to only multiply four such expressions (where sign next to $\sqrt{6}$ is a product of signs next to $\sqrt{2}$ and $\sqrt{3}$.): $$(a+b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a+b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) = a^2 + 2b^2 + 3c^2 + 6d^2 + (2ab - 6cd)\sqrt{2}$$ $$(a-b\sqrt{2} + c\sqrt{3} - d\sqrt{6})(a-b\sqrt{2} - c\sqrt{3} + d\sqrt{6}) = a^2 + 2b^2 + 3c^2 + 6d^2 - (2ab - 6cd)\sqrt{2}$$

So, product of all four expressions is $(a^2 + 2b^2 + 3c^2 + 6d^2)^2 - 2(2ab - 6cd)^2.$

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