Since $(1+\sqrt2)^2=3+2\sqrt2$, the positive square root is $1+\sqrt2$ and generates $\mathbb Q(\sqrt2)$, a real quadratic field. A primitive cube root ω is nonreal and satisfies $x^2+x+1=0$, so it is not in that real field and adds degree two. By the tower law, [K:Q] = 2×2 = 4.
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