<s>[INST] Show that $x^4-10x^2+1$ is reducible over $\mathbb{Q}(\sqrt{6}), \mathbb{Q}(\sqrt{2}),$ and $\mathbb{Q}(\sqrt{3})$. Is there an easier method? Ok, I actually worked this out as I was typing it up. But my solution seems kind of inelegant and involves a lot of tedious algebra that I've omitted here. Can anyone think of an easier method?
(1) Reducibility over $\mathbb{Q}(\sqrt{6})$ 
Quite straightforward, as the substitution $y = x^2$ and the quadratic formula on $y^2 - 10y + 1$ give:
$x^4-10x^2+1 = (x^2-5-2\sqrt{6})(x^2-5+2\sqrt{6})$.
(2) Reducibility over $\mathbb{Q}(\sqrt{2})$
Further reduction gives: $(x-\sqrt{5-2\sqrt{6}})(x+\sqrt{5-2\sqrt{6}})(x-\sqrt{5+2\sqrt{6}})(x+\sqrt{5+2\sqrt{6}})$.
It's also not that hard to show $\sqrt{5-2\sqrt{6}} = \sqrt{2}-\sqrt{3}$ and 
 $\sqrt{5+2\sqrt{6}} = \sqrt{2}+\sqrt{3}$ so the reduction becomes:
$(x-\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})$.
If we label these a, b, c, d left to right then:
$a.c = x^2-2\sqrt{2}x -1$ and $b.d = x^2+2\sqrt{2}x -1$ so it's reducible over $\mathbb{Q}(\sqrt{2})$.
(2) Reducibility over $\mathbb{Q}(\sqrt{3})$
$a.d = x^2-2\sqrt{3}x+1$ and $b.c = x^2+2\sqrt{3} +1$ so it's reducible over $\mathbb{Q}(\sqrt{3})$.
 [/INST] Well, a possibility is to write it as a difference of two squares and thus factors into a sum and difference.  For e.g.: 
$$x^4-10x^2+1 = (x^2-5)^2 - 24$$
and noting that $24=4\cdot6$ is a square in $\mathbb{Q}(\sqrt{6})$, so this is the difference of two squares.
Similarly, $x^4-10x^2+1 = (x^2-1)^2 - 8x^2$ and note $8$ is a square in $\mathbb{Q}(\sqrt{2})$.  
Again, $x^4 - 10x^2 + 1 = (x^2+1)^2 - 12x^2$ and $12$ is a square in $\mathbb{Q}(\sqrt{3})$.
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