Splitting it every which Way

1.- Let the polynomial (x - 1)(x - 2)(x - 3) then $\mathbb{Q}(1,2,3) = \mathbb{Q} \Rightarrow [\mathbb{Q}: \mathbb{Q}] = 1$ . $\mathbb{Q}(a,b,c)$ is the smallest field containing $\mathbb{Q}$ and a, b and c. Notice also that $\mathbb{Q}(a,b,c) = \mathbb{Q}(a,b)$ due to Vieta's formula $a + b + c \in \mathbb{Z}$

2.- Now, let the polynomial $x^3 + x$ then $\mathbb{Q}(i,-i,0) = \mathbb{Q}(i) \Rightarrow [\mathbb{Q}(i) : \mathbb{Q}] = = 2$ .

3.-Let the polynomial $p(x) = x^3 - 3x + 1$ then if $\theta$ is a root of $p(x)$ , $\theta^2 - 2$ is also a root of $p(x)$ and $\mathbb{Q}(\theta,\theta^2 -2) = \mathbb{Q}(\theta) \Rightarrow [\mathbb{Q}(\theta) : \mathbb{Q}] = 3$ . Let's see that $p(x)$ is irreducible over $\mathbb{Q}$ . If some root $\theta$ of $p(x)$ belonged to $\mathbb{Q}$ then $\theta = \frac{m}{n}$ with m,n coprime integers and $\theta$ distinct to 1 and -1, then $\frac{m^3}{n^3} = -1 + 3\frac{m}{n} \Rightarrow m^3 = n^2 \cdot (3m - n)$ but this a contradiction with the Fundamental Arithmetic Theorem.

4.-And finally, the polynomial $x^3 - 2$ fullfils $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \cdot e^{\frac{2i \cdot \pi}{3}}) : \mathbb{Q}] =$ $= [\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \cdot e^{\frac{2i \cdot \pi}{3}}) : \mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 2 \times 3 = 6$ .

Note.- there aren't more cases. Why?