<s>[INST] 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?
 [/INST] I thought I'd get rid of the surds one-by-one: first get rid of $\sqrt2$, then $\sqrt3$, and finally $\sqrt5$. To get rid of a surd, I would get everything with that surd in it to one side, and then square. (Note that $\sqrt{15}$ has a $\sqrt3$ in it.)
\begin{align}
x&=\sqrt2+\sqrt3+\sqrt5\\
x-\sqrt3-\sqrt5&=\sqrt2\\
(x-\sqrt3-\sqrt5)^2&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+8)&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+6)&=0\\
x^2-2\sqrt5x+6&=2\sqrt3x-2\sqrt{15}\\
(x^2-2\sqrt5x+6)^2&=(2\sqrt3x-2\sqrt{15})^2\\
x^4-4\sqrt5x^3+32x^2-24\sqrt5x+36&=12x^2-24\sqrt5x+60\\
x^4-4\sqrt5x^3+20x^2-24&=0\\
x^4+20x^2-24&=4\sqrt5x^3\\
(x^4+20x^2-24)^2&=80x^6\\
x^8+40x^6+352x^4-960x^2+576&=80x^6\\
x^8-40x^6+352x^4-960x^2+576&=0
\end{align}
This, of course, is equal to the Galois theory answer of $\displaystyle\prod(x\pm\sqrt2\pm\sqrt3\pm\sqrt5)$, where you multiply every possible choice of signs together.
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