(This is a question from an engineer and extremely naïve mathematician when it comes to the topic of irrational and transcendental numbers and the precise distinction between them.)
I begin with two examples of what I mean by the title question:
Let $A_1=(a+\sqrt{b})$ where a and b are rational numbers. Find $A_2$ so that the product $A_1 A_2$ is rational?
Obviously the answer is $A_2=(a-\sqrt{b})$ with $A_1 A_2=a^2-b$
Another slightly more difficult example with $a$, $b$ and $c$ rational is $A_1=(a+\sqrt{b}+\sqrt[3]{c}\,)$
with a little more work I found that one possible value for $A_2$ is
$$A_2=\left(\sqrt[3]{c} \left(-\left(a+\sqrt{b}\right)\right)+\left(a+\sqrt{b}\right)^2+c^{2/3}\right) \left(a^3-\sqrt{b} \left(3 a^2+b\right)+3 a b+c\right),$$ with $A_1 A_2=a^6-3 a^4 b+2 a^3 c+3 a^2 b^2+6 a b c-b^3+c^2$
Generalising further: Is it possible to prove that for every algebraic irrational number multiplier (real or complex) (or at least for a large subset of them), there exists at least one non-equal finite closed form multiplicand, so the product of the two is a rational number?
Implying that the inverse $\frac{1}{A_1}=\frac{A_2}{q}$ can be "sensibly" calculated in closed form (with $q$ rational). I am using the word "sensibly" to signify that I have no clear and precise mathematical definition for exact nature and productive constraints on this closed form, just the examples above.
The underlying reason I ask is that it is not immediately clear to me that an irrational (possibly transcendental?) number like $(\zeta(3)+a)$ or $(\zeta(3)+\sqrt{a})$, where $a$ is rational, has an associated non-equal closed form multiplicand, with the product of the two, resulting in a rational number.