Operator ordering (for the Groenewold-van Hove theorem) Suppose we have the (un)usual Schrödinger representation $\pi'(\cdot)$ of the Heisenberg algebra, along with the extension in $\mathbb{sl}(2,\mathbb{R})$ for the quadratic polynomials. It is assumed that
$$
\pi'(q^3) = -iQ^3;\quad \pi'(p^3) = -iP^3,
$$
where $[Q,P] = i\mathbb{I}$ (the usual commutator in quantum mechanics). I want to prove that
$$
\pi'(q^2 p) = -\frac{i}{2}(Q^2P + PQ^2)\quad (1)
$$
using the Poisson bracket $\lbrace q^3,p^2\rbrace = 6q^2p$.
Is (1) correct? The most that I can do is, using that $\pi'$ is a linear map
$$
\begin{split}
\pi'(\frac{1}{6}\lbrace q^3,p^2 \rbrace) &= \frac{1}{6}[\pi'(q^3),\pi'(p^2)] \\
&= -\frac{1}{6}[Q^3,P^2]\\
&= -\frac{i}{6}(3QP^2 + 3P^2Q + 2QPQ),
\end{split}
$$
but I cannot get rid of the term $2QPQ$. Or is it that my commutator computation is wrong?
 A: The last line of your commutator evaluation is definitely wrong.
It appears you are striving to calculate the Groenewold anomaly but I have failed to understand your bizarre "linear"  map of "the usual Schrödinger representation ′(⋅)" you posit to antihermitean operators, so I'll avoid it completely.
I'll just evaluate a few commutators correctly for you:
$$
[Q^3,P^2]= 3i ~(PQ^2 + Q^2 P),\\
[Q^2,P^3]= 3i ~(QP^2 + P^2 Q), ~~\leadsto \\
[[Q^3,P^2], [Q^2,P^3]]=   12[Q^3,P^3] +\mathbf{36}~.
$$
The last (boldfaced), constant, term is the celebrated Groenewold anomaly (has an $\hbar^2$ factor when you reinstate ℏ in your Heisenberg commutation relation).
Virtually  straightforward to compute in deformation quantization.
It corrects the classical limit expression
$$
\{\{q^3,p^2\}, \{q^2,p^3\}\}=   12\{q^3,p^3\} ,
$$
which has lost quantum information, as usual.

Edit post comments
Having found a copy of Woit's book, I at last appreciated the sadistic notation. His conceit (so 1930s) is to define a trial quantization map $i\pi'(q,p)$, Hermitian,  for the unambiguous low polynomial orders, and then attempt to ladder his way up to higher orders defining them through the chimeric desideratum
$$
\pi'(\{ f,g\})= [\pi'(f),\pi'(g)], 
$$
producing
$$
i\pi'(qp)=i\pi'(\{q^2,p^2\}/4)= -\frac{i}{4}[Q^2,P^2]=\tfrac{1}{2}(PQ+QP), 
$$(in Weyl ordering), and further producing your (1), which, however, is not in Weyl ordering! (You simply mishandled the commutator).
He then concludes his chapter by pointing out the hapless chimeric desideratum above is inconsistent, as the quartic term $\pi'(q^2p^2)$ is ill-defined, as evaluated in two alternate PB pathways. Mathematicians have these circuitous dandy arguments, but the point is no consistent  map can be found. Above, I condensed Groenewold's direct argument in one line, just as in our booklet.
