Let $F = \mathbb{Q}$ and let $f(x) \in F[x]$ be defined as $f(x) = x^3 + 7x + 101$ (irreducible). Let $\alpha, \beta, \gamma$ be algebraic numbers such that $f(\alpha) = f(\beta) = f(\gamma) = 0$, but $\alpha, \beta, \gamma$ are not necessarily distinct. Now let $K = F(\alpha, \beta, \gamma)$.
Claim 1. $[K:F] = 27$; i.e. the numbers $\alpha^a \beta^b \gamma^c$ are linearly independent for $a,b,c \in \{0,1,2\}$.
Claim 2. $|\text{Gal}(K/F)| = 6$; i.e. the set of $F$-automorphisms of $K$ are precisely those automorphisms which permute $\{\alpha,\beta,\gamma\}$, hence the number of such automorphisms is $3! = 6$.
Theorem (from textbook). $|\text{Gal}(K/F)|$ divides $[K:F]$.
Contradiction. In fact, $6$ does not divide $27$.
What went wrong here?
The numbers $\alpha^a\beta^b\gamma^c$ are not necessarily linearly independent for $a,b,c\in \{0,1,2\}$.
For $F(\alpha,\beta,\gamma)$, a basis of this should be $\alpha^a\beta^b\gamma^c$ where $1 \le c \le [F(\gamma):F]$, $1 \le b \le [F(\gamma,\beta):F(\gamma)]$ and $1 \le a \le [F(\alpha,\beta,\gamma):F(\beta,\gamma)]$. So your counting of $27$ is not quite correct. In fact, over $F(\gamma)$, we have $x^3+7x+101=(x-\gamma)(x^2+mx+n)$ so $\beta$ is root of polynomial $x^2+mx+n$ so $[F(\beta,\gamma):F(\gamma)]\le 2$.