Supposing you want to define an algebraic number as the root of a polynomial. The problem is the polynomial has more than one root. But each root is has different properties. (Or do they?)
If the roots are from irreducible polynomials, e.g. roots of $x^3-2=0$, does any operations in the field depend on the particular root you choose?
i.e. the polynomial $(x-3)(x^2-3)=0$ has two types of roots.
But in general if you multiplied two algebraic numbers which is the nth and mth root of polynomials. Could you say the result is an algebraic number which is the root of a bigger polynomial which is the pth root.
Which field? There are several relevant fields you may be having in mind.
If $F$ is a field and $p(x) \in F[x]$ an irreducible polynomial, the field $K=F[x]/\langle p(x) \rangle$ is an extension field of $F$ containing a root of $p$. There's also the splitting field $L$ of $F$ which extends $F$ and contains all the roots of $p$. And finally, when the base field $F$ is the rationals $\mathbb{Q}$, each of $K$ and $L$ can be realized as subfields of $\mathbb{C}$; in the case of $K$, this may be doable in more than one way, as is the case with your example $p(x)=x^3-2$.
The different embeddings of $K$ in $\mathbb{C}$ are all isomorphic, so no operations inside those fields "depend" in any way on which embedding you've chosen (meaning, which root of the complex roots you've chosen to bring in). The abstract field $K$ of course doesn't depend on any such choice, since it has a unique definition independent of any choices.
The field $L$ is different (in your case, it has degree $6$ over $\mathbb{Q}$, instead of $3$), but it, too, is uniquely determined. No choices involved.
The product of two algebraic numbers is an algebraic number, yes. There's no natural order on the roots, if that's what you mean, so we can't make a statement like "the 3rd root of this polynomial times the 5th root of that polynomial is the 12th root of this other polynomial."