Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let 
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial.  Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?

I started out by observing that $2+3 = 5$; I thought that might make it easier.  Then
$$
x-\sqrt{5} = \sqrt{2}+\sqrt{3}
$$
$$
x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5
$$
$$
x^2-2\sqrt{5}x = 2\sqrt{6}
$$
$$
x^4-4\sqrt{5}x^3+20x^2 = 24
$$
$$
x^4+20x^2-24 = 4\sqrt{5}x^3
$$
$$
x^8+40x^6+352x^4-960x^2+576 = 80x^6
$$
$$
x^8-40x^6+352x^4-960x^2+576 = 0
$$
(Or, of course, you can do it more systematically via the Galois theory!)
