Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below

Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.

All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ and $2^{1/3}\omega^2$, where $\omega = (-1+i\sqrt{3})/2$ is a cube root of unity. Thus $\mathbb{Q}(2^{1/3},\omega)$ or $\mathbb{Q}(2^{1/3},i\sqrt{3})$ is a splitting field of $f(x)$. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ as irreducble polynomial of $2^{1/3}$ is $x^3-2$(Eisenstien's criteria for $p=2$) and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$, as irreducible polynomial of $\omega$ is $g(x) = x^2+x+1$( $g(x+1)$ is irreducble by eisenstien's criteria for $p=3$). Therefore, if we have a splitting field $\mathbb{Q}(z)$, then degree of irreducible polynomial of $z$ is $6$ by compositum of fields. Now we need to find $z$.
Me and my friend got the idea of taking $z$ as $2^{1/3} + i\sqrt{3}$.
$$\begin{eqnarray*} x & = &  2^{1/3} + i\sqrt{3} \\ \left( x - 2^{1/3} \right)^2 & = & -3 
\\ -2^{2/3}x+2^{2/3} & = &  -x^2 -3 \\ 4(x-1)^3 & = & (x^3+3)^3 \end{eqnarray*}$$
Thus $h(x) = (x^2+3)^3-4(x-1)^3$ is a degree six polynomial having root $2^{1/3} + i\sqrt{3}$. How do i show from here that it is irreducible in $Q[x]$? It's expanded form is $x^6+9x^4-4x^3+39x^2-12x+31$.

Actually, the polynomial is irreducible over $\Bbb F_{19}$ and hence irreducible over $\Bbb Z$ and $\Bbb Q$. It has no root over $\Bbb F_{19}$ and no quadratic or cubic factor.
