Index Manipulation and Angular Momentum Commutator Relations I have been trying for hours and cannot figure it out. I am not asking anyone to do it for me, but to understand how to proceed. 
We have the relations 
$$[L_i,p_j] ~=~ i\hbar\;  \epsilon_{ijk}p_k,$$
$$[L_i,r_j] ~=~ i\hbar\;  \epsilon_{ijk}r_k,$$
$$[L_i,L_j] ~=~ i\hbar\;  \epsilon_{ijk}L_k.$$
Now I am trying to calculate 
$$[L_i,(p\times L)_j].$$ 
I do not know how to reduce it. When I try I am left with too many dummy indices that I don't know what to do. For example, using 
$$[AB,C] = A[B,C]+[A,C]B,$$ 
and expanding the cross product in terms of Levi-Civita symbols 
$$[L_i,\epsilon_{jmn}p_m L_n],$$
but I don't know how to proceed correctly from here. For example, I tried, 
$$ =~ \epsilon_{jmn}(\;p_m[L_i,L_n]+[L_i,p_m]L_n ).$$ 
Is this correct? If so, in the next step I used the known commutation relations
$$ =~i\hbar\; \epsilon_{jmn}(\; \epsilon_{ink}p_mL_k+\epsilon_{imk}p_kL_n).$$ 
Once again, I am stuck and do not know how to evaluate this further. Could someone tell me what I am doing wrong, or if not, how to proceed? 
 A: You should use the Levi-Civita reduction formula
$$\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$
and using the fact that $(a\times b)_i=\epsilon_{ijk}a_jb_k$ you should be done.
A: It is possible to continue OP's calculation as follows
$$ =~i\hbar\; \epsilon_{jmn}(\epsilon_{ink}p_m L_k+\epsilon_{imk}p_k L_n)
~=~-i\hbar( \epsilon_{jkm} \epsilon_{min}+\epsilon_{kim} \epsilon_{mjn})p_k L_n $$ 
$$ \stackrel{\rm Jac.Id.}{=}~i\hbar\; \epsilon_{ijm} \epsilon_{mkn}p_k L_n 
~=~i\hbar\; \epsilon_{ijm}(p\times L)_m ,$$ 
where one uses the Jacobi identity 
$$\sum_{{\rm cycl.}~i,j,k} \epsilon_{ijm} \epsilon_{mkn}~=~0 $$
for the Levi-Civita symbols.
A: Let's see:
$\textbf{L}=\textbf{r}\times \textbf{p}$
$\textbf{p}\times\textbf{L}=\textbf{p}\times (\textbf{r}\times \textbf{p})=\textbf{p}^2 \textbf{r}-(\textbf{p}\cdot \textbf{r})\textbf{p}$
The second  term is not zero ($\textbf{p}\cdot \textbf{r}\neq 0$) as in classical mechanics because of $[x_j,p_j]=i\hbar$. Therefore
$[L_i,(\textbf{p}\times\textbf{L})_j]=[L_i,\textbf{p}^2r_j]-[L_i,(\textbf{p}\cdot \textbf{r}) p_j]$
However
$[L_i,\textbf{p}^2r_j]=\textbf{p}^2[L_i,r_j]=i\hbar \textbf{p}^2\epsilon_{ijk}r_k$ 
because $[L_i,\textbf{p}^2]=[L_i,p_m p_m]=[L_i,p_m]p_m+p_m[L_i,p_m]=i\hbar\epsilon_{ijk}(p_k p_m+p_m p_k)=0$ 
and 
$[L_i,(\textbf{p}\cdot \textbf{r}) p_j]=[L_i,p_m r_m p_j]=[L_i,p_m]r_m p_j+p_m[L_i,r_m]p_j+p_m r_m[L_i,p_j]$
but $[L_i,p_m]r_m p_j+p_m[L_i,r_m]p_j=i\hbar\epsilon_{iml}p_l r_m p_j+i\hbar\epsilon_{iml} p_m r_l p_j=0$
and $p_m r_m[L_i,p_j]=i\hbar(\textbf{p}\cdot \textbf{r})\epsilon_{ijk}p_k$
Therefore 
$[L_i,(\textbf{p}\times\textbf{L})_j]=i\hbar \textbf{p}^2\epsilon_{ijk}r_k-i\hbar(\textbf{p}\cdot \textbf{r})\epsilon_{ijk}p_k=i\hbar\epsilon_{ijk}[\textbf{p}^2r_k-(\textbf{p}\cdot \textbf{r})p_k]=i\hbar\epsilon_{ijk}(p\times L)_k$
This is true for any $A_j$: $[L_i,A_j]=i\hbar\epsilon_{ijk}A_k$
