Why does Drinfeld Unitarization work? In Drinfeld's paper "Quasi-Hopf Algebras" he illuminates a process by which you can replace the $R \in A \otimes A$ associated to a quasi-Hopf QUE-algebra $(A, \Delta, \varepsilon, \Phi)$ over $k[[h]]$ with an $\overline{R}$ which satisfies the unitary condition. I describe the process here:
$\overline{R} = R*(R^{21}*R)^{-\frac{1}{2}}$
Using Sweedler notation  if $R = \Sigma (r_1 \otimes r_2)$ then $R^{21} := \Sigma(r_2 \otimes r_1)$.  The unitary condition states $RR^{21} = 1$. His argument for why $\overline{R}$ satisfies this condition is as follows:
$\overline{R}^{21}\overline{R} = R^{21}*(R*R^{21})^{-\frac{1}{2}}*R*(R^{21}*R)^{-\frac{1}{2}} = R^{21}*(R*R^{21})^{-1}*R = 1$
My question is about this second equality here.  How does he arrive at this?  Keep in mind that $A$ is not a commutative algebra. If there are any clarification questions please ask!  Thank you so much!
 A: Indeed those papers of Drinfeld are very terse.  [My most sucessful use of MathOverflow] was also an explanation of an offhand remark in such a paper.
I think that the best reference for this construction is the paper by Berenstein-Zwicknagl https://arxiv.org/abs/math/0504155, see especially the later half of section 1.
In this paper, they describe $ \bar R $ in the case where $ A$ is a quantum group (I mean the quantized universal envelopping algebra of a semisimple Lie algebra). Pick two representations $ V, W $ that you want to braid.  Consider the action of $ R $ on $ V \otimes W $.  All of the eigenvalues of $ R $ will be of the form $ \pm q^k $ for some $ k \in \mathbb Z$.  Then this factor $ (R^{21} R)^{-1/2} $ is a diagonal matrix which gets rid of these powers of $ q $.
Hope that helps!
A: The short answer to your question is that if $x,y$ are elements in an algebra in topologically free $k[[\hbar]]$-modules whose constant term is 1, then they have a unique square root whose constant term is also one, and if $x,y$ commute then say the square root of $x$ also commutes with $y$. Indeed if $a$ is the square root of $x$, then
$$(yay^{-1})^2=ya^2y^{-1}=x$$
and because $yay^{-1}$ also has constant term equal to 1, we get $yay^{-1}=a$. This shows at once that $(RR^{2,1})^{\frac12}$ commutes with $R$.
One way to think about it is as follow (this is also explained in Joel's paper). Any finitely generated group $G$ has a so-called pro-unipotent aka malcev aka rational completion $G(\mathbb{Q})$. One of its definition is that it is the univrsal uniquely divisible group having a morphism from $G$. In other words, it is the universal group in which images of elements of $G$ have a unique $n$th root for any $n$. So roughly elements of this groups are the $x^{\lambda}$ where $x \in G$ and $\lambda \in \mathbb{Q}$. Now the same argument as above shows if $x,y$ commute, then so do any possibly rational power of their image in $G(\mathbb{Q})$ (uniqueness is again key here).
This has a relative version, where in the case at hand you roughly speaking apply this construction to the pure braid group $P_n$ inside of the braid group $B_n$: you get a certain group $B_n(\mathbb{Q})^{rel}$ fitting into an exact sequence
$$1 \rightarrow P_n(\mathbb{Q}) \rightarrow B_n(\mathbb{Q})^{rel} \rightarrow S_n\rightarrow 1. $$
Long story short you get this way a morphism from the so-called cactus group $\Gamma_n$ (the group of which coboundary categories give representations) into $B_n(\mathbb{Q})^{rel}$ by taking square roots of the generators of $P_n$ inside there. Now for any quantized quasi-Hopf algebra, of more generally in any braided tensor category over $k[[\hbar]]$ in which the braiding satisfies
$$\beta_{U,V}\beta_{V,U} =id_{U\otimes V} +O(\hbar)$$
the representations of $B_n$ you get factor through $B_n(\mathbb{Q})^{rel}$, hence restrict to representations of $\Gamma_n$.
