= Solution
In the position representation, $\hat p=-i\hbar\nabla$. The <orbital angular momentum> operator and its squared magnitude are
$$
\boxed{\hat L=\hat x\times\hat p=-i\hbar\hat x\times\nabla,\qquad
\hat L^2=\sum_{i=1}^3\hat L_i^2.}
$$
For example, with $D=x\cdot\nabla$, the latter is $-\hbar^2[r^2\Delta-D^2-D]$. Operator products here act on smooth <wavefunctions> in a common invariant domain.
To calculate the <angular momentum commutation relations>, use $[x_a,p_b]=i\hbar\delta_{ab}$ and the vanishing position-position and momentum-momentum <commutators>. Expanding the products gives
$$
[x_ap_b,x_cp_d]=i\hbar(\delta_{ad}x_cp_b-\delta_{bc}x_ap_d).
$$
Multiplication by $\epsilon_{iab}\epsilon_{jcd}$ then gives
$$
\begin{aligned}
[L_i,L_j]
&=i\hbar\epsilon_{iab}\epsilon_{jcd}
(\delta_{ad}x_cp_b-\delta_{bc}x_ap_d)\\
&=i\hbar(x_i p_j-x_j p_i)
=i\hbar\epsilon_{ijk}L_k.
\end{aligned}
$$
Finally, the product commutator rule yields
$$
[L^2,L_i]=\sum_j\bigl(L_j[L_j,L_i]+[L_j,L_i]L_j\bigr)
=i\hbar\sum_{j,k}\epsilon_{jik}(L_jL_k+L_kL_j)=0.
$$
The final cancellation pairs the antisymmetric <Levi-Civita symbol> with an expression symmetric in $j,k$. Thus
$$
\boxed{[L_i,L_j]=i\hbar\epsilon_{ijk}L_k,\qquad [L^2,L_i]=0.}
$$
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