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$\mathbb{Q}(\sqrt 6, \sqrt 10, \sqrt 15):\mathbb{Q}=\mathbb{Q}(\sqrt 6+ \sqrt 10+\sqrt 15):\mathbb{Q}$

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Is there a question? And how is this essentially different from math.stackexchange.com/questions/29846/simple-extension ? –  Arturo Magidin Mar 30 '11 at 18:17
Note that $\sqrt{6}\sqrt{10} = \sqrt{60} = 2\sqrt{15},$ and thus that $\mathbb Q(\sqrt{6},\sqrt{10},\sqrt{15}) = \mathbb Q(\sqrt{6},\sqrt{10})$. This might make this field easier for you to handle. –  Matt E Mar 30 '11 at 19:33

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HINT $\: $ Note $\ (\sqrt{2}+\sqrt{3}+\sqrt{5})^2\ =\ 2\ (\sqrt{6}+\sqrt{10}+\sqrt{15})\ +\ 10\:.\: $ Now refer to your prior question.

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