Yes. In fact, the strong form of Banach-Tarski paradox is as follows:
Theorem. Let $n\geq3$ and let $A,B\subseteq\mathbb{R}^n$ be arbitrary bounded sets with non-empty interior. Then $A$ and $B$ are equidecomposable.
This means that we can cut any such $A$ into finitely many pieces and using isometries (distance-preserving maps; i.e. linear maps & translations) put them back together to form $B$.
Note, that equidecomposability is easily seen to be an equivalence relation, from which it already follows that: if we can cut one ball into finitely many pieces and create two, we can also reverse the process.
To see that equidecomposability is a symmetric relation, just note that isometries of any metric space (including $\mathbb R^n$) form a group. So if you have decomposed $A$ into sets $A_1, A_2, ..., A_n$ and used isometries $g_1, g_2, ..., g_n$ to get sets $B_1, B_2,...,B_n$ which together form $B$ (we usually write $g_iA_i=B_i$ to denote the isometry $g_i$ is acting on a set $A_i$, i.e. $g_iA_i$ is the set we get from $A_i$ when we use the isometry $g_i$ on it), you can just use the inverses of these isometries to arrive from $B$ back to $A$. (So in the notation from above: if $g_iA_i = B_i$ for $i=1,2,...,n$, then we also have $A_i=g_i^{-1}B_i$. Put simply: if I have moved a set and rotated it a bit, I can always rotate and move it back, to get the original set back.)