Here’s how you can show irreducibility of your polynomial, using the same method I used in the argument I gave in the reference that @BillDubuque mentioned:
First form $x^3-5$, the minimal polynomial for $\root3\of5$ over $\Bbb Q$. Now think of it as a polynomial over the principal ideal domain $\Bbb Z[\sqrt2\,]$. It happens that $5$ is still prime there, so Eisenstein still applies, and this is the minimal polynomial for $\root3\of5$ over the extension. Consequently $(x-\sqrt2)^3-5$ is still irreducible over $\Bbb Z[\sqrt2\,]$, and it’s the minimal polynomial for $\sqrt2+\root3\of5$. Let’s call this polynomial $f(x)$. It’s actually $x^3-3\sqrt2x^2+6x-5+2\sqrt2$. Now take the conjugate polynomial, call it $\bar f$, which you get by replacing $\sqrt2$ by $-\sqrt2$ in $f$ wherever it appears. Finally, multiply, and get $f\bar f$, which turns out to be exactly the sextic polynomial you calculated.
This clearly is a $\Bbb Q$-polynomial that has $\sqrt2+\root3\of5$ as a root. But it is also a $\Bbb Z[\sqrt2\,]$-polynomial, and we know its factorization into irreducibles there, namely $f\bar f$. And by uniqueness of factorization there, this is the only possible factorization with coefficients in $\Bbb Z[\sqrt2\,]$. But a $\Bbb Q$-factorization of your polynomial would also be a $\Bbb Z[\sqrt2\,]$-factorization, and we already have one such, and it’s unique. So there is no factorization of your polynomial over $\Bbb Z$ or $\Bbb Q$.
Another way to do this is to interpret $\alpha$ as a $\mathbb{Q}$-linear map from $\mathbb{Q}(\sqrt[3]{2})$ to itself. With respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, this is represented by the matrix $$\begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}.$$ The characteristic and minimal polynomial is $$t^3 - (0+0+0) t^2 + 3 \cdot \mathrm{det} \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} t - \mathrm{det} \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= t^3 - 6t - 6.$$
Best Answer
Another method to show that $p(x)$ is irreducible over $\Bbb Q$ is to show that it is already irreducible over $\Bbb F_7$. This is easier to show.