Commutation of Angular and Linear Momentum

[atomicpedals]

If I have a relation such as $$[L_{j} , \vec{p}^2]=0$$ where j=x,y,z.

Can I re-write it as $$[L_{j}, \vec{p} \vec{p}]=0$$ and then evaluate it as though it were an identity? e.g. $$[A,BC]=[A,B]C+[B,A]C=...$$

 

[mathfeel]

If I have a relation such as $$[L_{j} , \vec{p}^2]=0$$ where j=x,y,z.

Can I re-write it as $$[L_{j}, \vec{p} \vec{p}]=0$$ and then evaluate it as though it were an identity? e.g. $$[A,BC]=[A,B]C+[B,A]C=...$$

If unsure, you are better off explicitly writing out the dot product:
$[ L_j, \vec{p}^2] = [L_j, \sum_{k} p_k^2] = \sum_{k} [L_j, p_k^2]$

Now you can apply the ABC rule. The one you quote is wrong, by the way. It should be:
$[A, BC] = B[A,C] + [A,B]C$

 

[atomicpedals]

Thanks for the help!