Of course. Let's start with the ##[l_2, l_3]## commutator
##
\begin{align*}
\langle m, l | [l_2^2, l_3^2]| l,m \rangle &= \langle m, l | l_2^2l_3^2 - l_3^2l_2^2| l,m \rangle \\
&= \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle \\
\end{align*}
##
Utilize the fact that ##l_i## are Hermitian and the eigenvalue equation ##l_3| l,m \rangle = m| l,m \rangle##, we get
##
\begin{align*}
\langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle &= m^2(\langle m, l | l_2^2| l,m \rangle - \langle m, l | l_2^2| l,m \rangle) \\
& = 0
\end{align*}
##
As for the commutator result, we have
##
\begin{align*}
[l_1^2, l_2^2] &= [l_1^2, L^2 - l_1^2 - l_3^2] \\
&= -[l_1^2, l_3^2] \\
& = [l_3^2, l_1^2],
\end{align*}
##
where we use that any component of angular momentum commutes with the total angular momentum. A similar proof can be done for the ##[l_2^2, l_3^2]## commutator.
Also, correct me if I am wrong but it appears that this is true for all ##|l,m\rangle##, not just the specific case of ##l=1##.
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