Equality of two quadratic position-momentum orderings ($\hat{x}{\hat{p}}^2\hat{x}=\hat{p}{\hat{x}}^2\hat{p}$)

On a common invariant domain for the operator products, let the position operator and momentum operator obey $[\hat{x},\hat{p}]=i\hslash I$. Then $\hat{x}\hat{p}=\hat{p}\hat{x}+i\hslash I$, so both displayed expressions equal $(\hat{p}\hat{x}{)}^2+i\hslash \hat{p}\hat{x}$. In position representation both act on a smooth function $f$ as $-{\hslash }^2(x^2{f}^{\prime \prime }+2x{f}^{\prime })$. This particular equality does not imply that arbitrary classical reorderings remain equivalent in quantum mechanics.