Let $a=\sqrt{2}$, and let $b=\sqrt{-3}$.
Let $\omega=\exp\left(i\left({\large{\frac{\pi}{3}}}\right)\right)$.
Noting that
- $\omega$ is a primitive $6$-th root of $1$.$\\[4pt]$
- $\sqrt[6]{8}=\sqrt{2}=a$.
it follows that the roots of $x^6-8$ in $\mathbb{C}$ are
$$a,a\omega,a\omega^2,a\omega^3,a\omega^4,a\omega^5$$
hence $K=\mathbb{Q}(a,\omega)$ is a splitting field of $f$.
Then since
$$
\omega
=
\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)
=
\frac{1}{2}+\frac{1}{2}i\sqrt{3}
=
\frac{1}{2}+\frac{1}{2}b
$$
it follows that we can also write $K=\mathbb{Q}(a,b)$, or explicitly, $K=\mathbb{Q}\left(\sqrt{2},\sqrt{-3}\right)$.
Note that $[\mathbb{Q}(a):\mathbb{Q}]=2$ and $[\mathbb{Q}(b):\mathbb{Q}]=2$.
Since $a\in\mathbb{R}$, and $b\not\in\mathbb{R}$, it follows that $[\mathbb{Q}(a,b):\mathbb{Q}(a)] > 1$.
From $[\mathbb{Q}(b):\mathbb{Q}]=2$, we get $[\mathbb{Q}(a,b):\mathbb{Q}(a)]\le 2$, hence $[\mathbb{Q}(a,b):\mathbb{Q}(a)]=2$
Hence, we get
$$
[K:\mathbb{Q}]
=
[\mathbb{Q}(a,b):\mathbb{Q}]
=
[\mathbb{Q}(a,b):\mathbb{Q(a)}]
{\,\cdot\,}
[\mathbb{Q}(a):\mathbb{Q}]
=
2{\,\cdot\,}2
=
4
$$
Best Answer
If we have any diamond of finite field extensions
$\hskip 3in$
where $M=LK$ and the primitive element theorem applies, then
$$\color{Blue}{[M:K]}\le\color{Red}{[L:F]}.$$
Proof. By PET let $L=F(\theta)$ where $\theta$ has minimal polynomial $m(x)\in F[x]$. Then $M=K(\theta)$. Say the minimal polynomial of $\theta$ over $K$ is $n(x)\in K[x]$. Since $m(x)\in F[x]\subseteq K[x]$ and $m(\theta)=0$ we must have $n(x)\mid m(x)$ and therefore $\deg n(x)\le \deg m(x)$ but $\deg n=[M:K]$, $\deg m=[L:F]$ and so the inequality is satisfied.
In particular you have $M=\Bbb Q(\sqrt[3]{2},\sqrt{-3})$, $K=\Bbb Q(\sqrt[3]{2})$, $L=\Bbb Q(\sqrt{-3})$, $F=\Bbb Q$, which implies the inequality $[\Bbb Q(\sqrt[3]{2},\sqrt{-3}):\Bbb Q(\sqrt[3]{2})]\le[\Bbb Q(\sqrt{-3}):\Bbb Q]=2$. Since $\Bbb Q(\sqrt[3]{2},\sqrt{-3})$ is the splitting field of $x^3-2$ whereas $\Bbb Q(\sqrt[3]{2})$ isn't, we have $\Bbb Q(\sqrt[3]{2},\sqrt{-3})\ne\Bbb Q(\sqrt[3]{2})$ hence the degree is $>1$. Since the degree is $>1$ and $\le2$ it must be $2$. This is the reasoning involved.