<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] An elementary way 
$(x-\sqrt2)^2=(\sqrt3+\sqrt5)^2 \implies x^2+2-2x\sqrt2=8+2\sqrt{15}\implies(x^2-6-2x\sqrt2)^2=((x^2-6)^2-4x(x^2-6)\sqrt2+8x^2)^2=60\implies((x^2-6)^2+8x^2-60)^2=(4x(x^2-6)\sqrt2)^2$ 
This resulting polynomial of $\mathbb{Q}[x]$ is obviously of degree 8
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