I have used it a few times and so far I have not seen an instance in which it did not give me the minimal polynomial over $\Bbb{Q}$.
The method is the following: For an element $a \notin \Bbb{Q}$ set $a=X$ and use arithmetic operations until we get an expression that looks like $p=0$ where $p \in \Bbb{Q}[X]$.
Example with $p=\sqrt2$:
$$X=\sqrt2 \iff X^2=2 \iff X^2-2=0$$
Example with $\sqrt3+\sqrt5$:
$$X=\sqrt3+\sqrt5 \iff X^2=8+2\sqrt{15}\iff (X^2-8)^2=60 \iff X^4 - 16 X^2 + 4=0$$
Example with $\sqrt3+ \sqrt[3]2$:
$$X=\sqrt3+ \sqrt[3]2 \iff (X-\sqrt3)^3=2 \iff X^3- 3 \sqrt3 X^2 + 9 X- 3 \sqrt3=2 \iff$$
$$(X^3 +9X -2)^2 = (3 \sqrt3 X^2+3 \sqrt3)^2 \iff$$
$$X^6 + 18 X^4 - 4 X^3 + 81 X^2 - 36 X + 4 = 27 X^4 + 54 X^2 + 27 \iff$$
$$X^6 - 9 X^4 - 4 X^3 + 27 X^2 - 36 X - 23=0$$
It is clear that the resulting polynomial will be normed and have $a$ as its root, but the last two examples are not obviously irreducible. Can I rely on this method to give me the minimal polynomials?
For example, suppose you set $X = \sqrt{3 + 2\sqrt{2}}$, and you try to find the minimal polynomial using $X^2 = 3 + 2\sqrt{2}$; $X^2 - 3 = 2\sqrt{2}$; $(X^2 - 3)^2 = 8 = X^4 - 6X^2 + 9$; $X^4 - 6X^2 + 1 = 0$. Your method would thus find $X^4 - 6X^2 + 1$ as the minimal polynomial.
However, secretly, $X$ is actually equal to $1 + \sqrt{2}$, which has minimal polynomial $X^2 - 2X - 1$. In fact, $X^4 - 6X^2 + 1$ is not irreducible over $\mathbb{Q}$, since $X^4 - 6X^2 + 1 = (X^2 - 1)^2 - 4X^2 = (X^2 - 2X - 1) (X^2 + 2X - 1)$.