Solving this momentum operator

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I would appreciate some help in understanding what how this operator is being multiplied in the following equation:

$$x_{op}p^2_{op,x}=\left(p_{op,x}x_{op}+i\hbar\right)p_{op,x}$$

(basically how to solve LHS to get RHS)

From the looks of it somehow there is a derivation applied to both $x$ and $p$ operator but I can't understand why as $x$ is before $p$ to start with and so the $d/dx$ in $p$ won't be applicable to $x$!?

In case you need the full solution it is from, it can be seen belowsolution

Many thanks!

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$$x_{op}p_{op,x}^2=x_{op}p_{op,x}p_{op,x}=(x_{op}p_{op,x})p_{op,x}$$ We know that $$[x_{op},p_{op,x}]=x_{op}p_{op,x}-p_{op,x}x_{op}=i\hbar$$ so we can move the second term on the other side $$x_{op}p_{op,x}=i\hbar+p_{op,x}x_{op}$$ Finally, we plug this result in the first equation, and you get your desired result