(I believe this belongs to computational algebraic number theory, but if additional/different tags are better, let me know.)
Consider the cubic equation $x^3-3x^2+x-3=0.$ We have that $x=3$ is a root. On the other hand, one can use one of the algorithms for solving cubics (the methods followed by Tartaglia or others), and find a different looking expression, namely $ x= 1 +\root 3\of{2+\frac{10}{3\sqrt3}} + \root 3\of{2-\frac{10}{3\sqrt3}}. $ In this example, one can check that the numbers are the same, by noting that $ \root3\of{2+\frac{10}{3\sqrt3}}=1+\frac1{\sqrt3}, $ but this leads to my question:
Suppose that $\alpha$ and $\beta$ are complex numbers with explicitly given expressions, and that $\alpha$ and $\beta$ belong to some finite extension of ${\mathbb Q}$.
By "explicitly given", one can mean that finitely many irreducible polynomials $p_i$ with integer coefficients are given, and that $\alpha$ and $\beta$ can be obtained from roots of these $p_i$ by radicals (as the expressions above), perhaps with additional information on which of the roots of the $p_i$ one is considering ("the smallest positive root," "the one whose imaginary part is positive and second largest," ...). If you see a more precise way of making sense of this, that is fine as well.
In the example I was discussing, if I had not noticed that the cubic root simplified as it did, a reasonable way of convincing myself the two numbers were actually the same would have been to compute them. Using lots and lots of digits.
Compute numerical approximations to $\alpha$ and $\beta$ using some reliable CAS. Can we explicitly find an a priori (and feasible?) bound for the number of digits one needs to compute to ensure that if both approximations coincide, then in fact $\alpha=\beta$?
The bound most likely would depend not just on the degree of the extension of ${\mathbb Q}$ where we can find $\alpha$, $\beta$, and all the radicals that make them up, but I imagine the degree is part of it.
(Somebody told me the LLL algorithm is perhaps the way of doing this, but this is all new to me.)