= Solution
Use $[AB,C]=A[B,C]+[A,C]B$ and the <canonical commutation relation>. With $L_i=\varepsilon_{iab}x_ap_b$,
$$
\begin{aligned}
[L_i,x_j]
&=\varepsilon_{iab}x_a[p_b,x_j]
=-i\hbar\varepsilon_{iaj}x_a
=i\hbar\varepsilon_{ijk}x_k,\\
[L_i,p_j]
&=\varepsilon_{iab}[x_a,p_j]p_b
=i\hbar\varepsilon_{ijb}p_b
=i\hbar\varepsilon_{ijk}p_k.
\end{aligned}
$$
Using these two transformation laws on $L_j=\varepsilon_{jab}x_ap_b$ gives
$$
\begin{aligned}
[L_i,L_j]
&=\varepsilon_{jab}([L_i,x_a]p_b+x_a[L_i,p_b])\\
&=\boxed{i\hbar\varepsilon_{ijk}L_k}.
\end{aligned}
$$
These are the <orbital angular momentum commutation relations>.
Now
$$
[L^2,L_i]=\sum_j\bigl(L_j[L_j,L_i]+[L_j,L_i]L_j).
$$
The factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:
$$
\boxed{[L^2,L_i]=0}.
$$
The same commutators show that $p^2=p_jp_j$ and $r^2=x_jx_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric $\varepsilon_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and for
$$
H=\frac{p^2}{2m}+U(r)
$$
we have $[H,L_i]=0$, hence
$$
\boxed{[H,L^2]=0}.
$$
In particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By <simultaneous diagonalization>, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the <rotational invariance of a central-potential Hamiltonian>.
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