Find polynomial whose root is sum of roots of $X^5 - 2$ and $X^4 - 3$

I have a task where I need to find a polynomial $R \in \mathbb{Q}[X]$ that has roots $\alpha + \beta$ where $\alpha$ are arbitrary roots of $X^5 - 2$ and $\beta$ of $X^4 - 3\enspace.$

The general technique to use is obvious to me, the polynomial is given by $$R = \prod_{i, j} (x - (\alpha_i + \beta_j))$$ with the corresponding roots \begin{align*} \alpha_1 &= \sqrt{2}\\ \alpha_2 &= -\sqrt{-2}\\ \alpha_3 &= (-1)^{\frac{2}{5}}\sqrt{2}\\ \alpha_4 &= -(-1)^{\frac{3}{5}}\sqrt{2}\\ \alpha_5 &= (-1)^{\frac{4}{5}}\sqrt{2}\\ \beta_1 &= \sqrt{3}\\ \beta_2 &= -\sqrt{3}\\ \beta_3 &= i\sqrt{3}\\ \beta_4 &= -i\sqrt{3}\enspace. \end{align*} (compare to questions like Find polynomial whose root is sum of roots of other polynomials)

Now I know that one can argue that $R$ is indeed in $\mathbb{Q}[X]$, but we didn't yet step into Galois-theory so I probably can't use that, nor do I know the details of how to prove this.

The problem with manually computing $R$ is that the resulting equation is huge, really huge. After some hours of Mathematica magic I could resolve the above to \begin{align*} S &= X^{20} - 15X^{16} -8X^{15} + 90X^{12} -1560X^{11} + 24X^{10} - 270X^8 - 11160X^7\\ &\qquad- 4080X^6 - 32X^5 + 405X^4 - 7560X^3 + 3960X^2 - 480X - 227\enspace. \end{align*}

The problem is that, if I go from this direction, I would need to check all $5 \cdot 4 = 20$ roots. But that is again a huge task.

At this point I'm stuck, both directions aren't eligible for a manual approach. I probably need some kind of fancy argument with which I can follow that $R = S$ or that $S$ has those roots (despite the fact that $S$ fall out of the sky).

Alternatively, arguments that $R$ is in $\mathbb{Q}[X]$ due to construction. But in this case I would need some guidance because, as said, I'm not sufficiently acquainted with Galois-theory.

I'm not sure if this this is the argument you want. To be precise, what I want to explain is that for monic irreducible polynomials $$f(x)$$ and $$g(x)$$, and their roots $$\alpha_i$$, $$\beta_j$$ are the conjugates of the roots of them, respectively, the polynomial $$p(x) = \prod_i \prod_j \left( x-\alpha_i -\beta_j \right)$$ is in $$\mathbb{Z}[x]$$ without referring the Galois theory.
Viewed as a polynomial over $$\mathbb{Z}[\alpha_1, \alpha_2, \ldots , \alpha_n]$$, the coefficients of $$p(x)$$ are symmetric polynomials of $$\beta_j$$s, say $$\sigma_k$$ be the degree $$k$$ elementary symmetric polynomial of $$\beta_j$$s. Therefore coefficients of $$p(x)$$ are of the form $$B \left( \sigma_1 , \sigma_2 , \ldots , \sigma_m , \alpha_1 , \alpha_2 , \ldots , \alpha_n \right)$$. Hence $$B$$ is a polynomial of $$\alpha_i$$s with integral coefficients. However, $$B$$ is also symmetric of $$\alpha_i$$. Therefore, is an integer. This proves that $$p(x)$$ is in $$\mathbb{Z}[x]$$.