Is $x^6-x^3+1$ irreducible over $\mathbb Q[x]$?

Question

Is $x^6-x^3+1$ irreducible over $\mathbb Q[x]$?

Approach

If $x^6-x^3+1$ is reducible over $\mathbb Q[x]$, then it can be factored out with degree $1,2\;\text{or}\;3$.

So check that $x^6-x^3+1$ has a root over $\mathbb Z_2[x]\;\text{and}\;\mathbb Z_3[x]$

Then it has no roots in $\mathbb Z_2[x]\;\text{and}\;\mathbb Z_3[x]$

For degree $1$, $(x-1)$ and $(x+1)$ are not possible cases.

Hence I conclude that $x^6-x^3+1$ is irreducible over $\mathbb Z[x]$

Thus $x^6-x^3+1$ is irreducible over $\mathbb Q[x]$

My Question is that the first assumption, if $x^6-x^3+1$ is irreducible over $\mathbb Q[x]$, then it can be factored out with degree $1,2\;\text{or}\;3$, is whether valid or not valid.

EDIT

$x^6-x^3+1$ has a root $2$ over $\mathbb Z_3[x]$

So, my attempts was false.

Answer 1

$\newcommand{\Z}{\mathbb{Z}}$It seems proving irreducibility of $x^6 - x^3 + 1$ in $\Z[x]$ can reduced to an application of Eisenstein's criterion by the substitution $x = y - 1$.

Answer 2

$$x^6-x^3+1 = \frac{x^9+1}{x^3+1} = \frac{(x^{18}-1)(x^3-1)}{(x^9-1)(x^6-1)} = \Phi_9(-x)=\Phi_{18}(x) $$ and every cyclotomic polynomial is irreducible over $\mathbb{Q}$: in this case, as already shown by Andreas Caranti, it is enough to replace $x$ with $y-1$ and apply Eisenstein's criterion. An alternative is given by checking that $x^6-x^3+1$ is irreducible over $\mathbb{F}_{29}$.