Subgroup generated by $1 - \sqrt{2}$, $2 - \sqrt{3}$, $\sqrt{3} - \sqrt{2}$ For a number field $K$, Dirichlet's unit theorem says that$$(O_K)^\times = \mathbb{Z}^{r - 1} \oplus (\text{a finite cyclic group}),$$where $r$ is the number of all infinite places of $K$. An infinite place of $K$ is either an embedding of $K$ (as a field) into $\mathbb{R}$ or the complex conjugacy class of an embedding of $K$ (as a field) into $\mathbb{C}$ with dense image.
When $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, how do I see that $1 - \sqrt{2}$, $2 - \sqrt{3}$, and $\sqrt{3} - \sqrt{2}$ generate a subgroup $\Gamma$ of $(O_K)^\times$ such that$$\Gamma \cong \mathbb{Z}^3 = \mathbb{Z}^{r - 1}?$$

Expanding what I believe Franz Lemmermeyer meant with his hint. This is closely built into all the other answers as well (+1 to y'all), but I learned a trick from reading all this, so I can't keep the lid on...
Assume contrariwise that there exist a non-trivial triple $(a,b,c)$ of integers such that
$$
|(1-\sqrt2)^a(2-\sqrt3)^b(\sqrt3-\sqrt2)^c|=1.\qquad(i)
$$
I use the absolute value signs so that I don't need to worry about the torsion unit. Let's apply the automorphism $\tau:\sqrt2\mapsto\sqrt2,\sqrt3\mapsto-\sqrt3$ to equation $(i)$. As $\tau(1-\sqrt2)=(1-\sqrt2)$, $\tau(2-\sqrt3)=(2-\sqrt3)^{-1}$ and $\tau(\sqrt3-\sqrt2)=-(\sqrt3-\sqrt2)^{-1}$, we get another equation
$$
|(1-\sqrt2)^a(2-\sqrt3)^{-b}(\sqrt3-\sqrt2)^{-c}|=1.\qquad(ii)
$$
Multiplying $(i)$ and $(ii)$ together gives
$$
|(1-\sqrt2)^{2a}|=1\qquad(iii)
$$
from which we can infer that $a=0$.
There are several ways to continue. By using another automorphism $\sigma:\sqrt2\mapsto-\sqrt2, \sqrt3\mapsto-\sqrt3$ we similarly reach the conclusion $c=0$. Continuing in the same vein, using the automorphism $\sigma\tau$ gives us $b=0$.
Essentially we are taking the relative norms of these units to the three quadratic subfields and checking what that gives us.
Alternatively, after having derived $(iii)$ we can use Mathmo123's idea. The equations $(i)$ and $(iii)$ imply that $(2-\sqrt3)^{2b}$ and $(\sqrt3-\sqrt2)^{2c}$ are inverses of each other. But the former is an element of $\Bbb{Q}(\sqrt3)$ and the latter of $\Bbb{Q}(\sqrt6)$. The intersection of the two fields is $\Bbb{Q}$, so both must be rational. But the only rational units are $\pm1$, so this is absurd.
