= Solution
The <orbital angular momentum> operators are
$$
\boxed{L_1=x_2p_3-x_3p_2,\qquad L_2=x_3p_1-x_1p_3,\qquad L_3=x_1p_2-x_2p_1.}
$$
On smooth wavefunctions, $[x_i,p_j]=i\hbar\delta_{ij}$, while position operators commute with each other and momentum operators commute with each other. Applying the product rule for <commutators> gives
$$
[x_ip_j,x_kp_l]=-i\hbar\delta_{jk}x_ip_l+i\hbar\delta_{il}x_kp_j.
$$
Use it on the four terms in $[L_1,L_2]$: only $[x_2p_3,x_3p_1]=-i\hbar x_2p_1$ and $[x_3p_2,x_1p_3]=i\hbar x_1p_2$ survive, with the latter entering positively from the two minus signs. Therefore
$$
\boxed{[L_1,L_2]=i\hbar L_3.}
$$
The cyclic versions similarly give $[L_3,L_1]=i\hbar L_2$ and $[L_3,L_2]=-i\hbar L_1$. Thus, for $L_\pm=L_1\pm iL_2$,
$$
\boxed{[L_3,L_\pm]=\pm\hbar L_\pm.}
$$
Now define $L^2=L_1^2+L_2^2+L_3^2$. For each $i$,
$$
[L^2,L_i]=i\hbar\sum_{j,k}\epsilon_{jik}(L_jL_k+L_kL_j)=0,
$$
because the <Levi-Civita symbol> is antisymmetric in $j,k$ and the operator pair is symmetric. Linearity then gives \b[$\boxed{[L^2,L_\pm]=0}$]. The PDF superscript here is the squared total angular momentum, not the component $L_2$.
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