Irreducibility test in a number field
It suffices to show that $g(x)=x^3 + 3 \sqrt[3]6\,x + 1$ is irreducible over $K=\mathbb{Q}(\sqrt[3]6)$ because $$[\mathbb{Q}(\alpha):K]=3 \implies [\mathbb{Q}(\alpha):\mathbb{Q}] = 9$$ There are general algorithms to factor polynomials over number fields. However, I present two ad hoc ways for this particular polynomial.
First method
If $g$ is reducible, then it has a root in $K$. The only real root of $g$ is $\gamma= -0.182329$, this number is an algebraic unit. Hence $-1/\gamma = 5.48461$ is also an algebraic unit.
Let $u>1$ be the fundamental unit of a real cubic field with one real embedding, then $$u^3 > \frac{|\delta|-27}{4}$$ where $\delta$ is the discriminant of the field.
An instructive proof is given in Number Fields by Daniel Marcus (chapter 5). Introductory algebraic number theory by Alaca (chapter 13) contains an excessively verbose proof.
Now apply this proposition to ring of integer of $K$, with $\delta = -972$, we found its fundamental unit is $>6.1819$. Therefore $-1/\gamma \notin K$ so $\gamma \notin K$.
Second method
Replace $x$ by $x/\sqrt[3]6$ in $g(x)$ gives $x^3+18x+6$, irreducible over $\mathbb{Q}$. Denote a real root of $x^3+18x+6$ by $\gamma$. It suffices to show $\gamma \notin K$. Assume $\gamma \in K$, then $\mathbb{Q}(\sqrt[3]6) = \mathbb{Q}(\gamma)$. Both $$\{1,\sqrt[3]6, \sqrt[3]6^2\} \qquad \{1,\gamma, \gamma^2\}$$ are basis of $K$ over $\mathbb{Q}$. Let $\text{tr}$ denote the trace of $K$ over $\mathbb{Q}$. There exists $a,b,c\in \mathbb{Q}$ such that $$\gamma = a + b\sqrt[3]6+c\sqrt[3]6^2$$ Taking trace both sides gives $$0=\text{tr}(\gamma) = 3a \implies a=0$$ So $$\gamma^2 = b^2\sqrt[3]6^2+c^2\sqrt[3]6^4+12bc$$, taking trace again gives $$\tag{1} -36=\text{tr}(\gamma^2) = 36bc \implies bc=-1$$ Consider $\gamma^3$ gives $$\tag{2}-18=\text{tr}(\gamma^3)=18(b^3+6c^3) \implies b^3+6c^3=-1$$ It is easy to see that $(1),(2)$ have no rational solutions, this contradiction shows $\gamma \notin K$.
This answer elaborates on the observations made by @JyrkiLahtonen in the comments. Let $\alpha$ be a root of $x^9+3x^6+165x^3+1$ then $$\tag1 (\alpha^3+1)^3+6(3\alpha)^3=0$$ and $$\tag2 3(\alpha^3+7)^3+2(\alpha^3-8)^3=0.$$ Equation $(1)$ is already mentioned in the question and $(2)$ was found by observing that $$(1-b)(\alpha^3+a)^3+(a-1)(\alpha^3+b)^3$$ is linear in $\alpha^3$ for any $a,b$ and then picking suitable values. From this it follows that $\mathbb{Q}(\alpha)$ contains both a cube root of $3$ and a cube root of $2$. As mentioned in the comments, this suffices to show that $\alpha$ has degree $9$ over $\mathbb{Q}$.