How to calculate $\left[ \vec{L}^2, x_i \right]$

I've been asked to prove $$\left[ \vec{L}^2, x_i \right] = -2i\hbar \varepsilon_{ijk}L_j x_k -2\hbar^2 x_i$$ and I don't seem to get it correctly.

I propose $$\left[ \vec{L}^2, x_i \right] = \left[ L_l L_l, x_i \right] = L_l\left[ L_l, x_i \right] +\left[ L_l, x_i \right]L_l$$ which leads me to $$\left[ \vec{L}^2, x_i \right] = -i\hbar \varepsilon_{ilk}L_lx_k - i\hbar\varepsilon_{ilk}x_kL_l$$ but I think I'm using some property not properly because I arrive to the following result $$\left[ \vec{L}^2, x_i \right] = -i\hbar \varepsilon_{ilk}L_lx_k -i\hbar\varepsilon_{ilk}\left[L_lx_k -i\hbar\varepsilon_{lkm}x_m\right] =-2i\hbar \varepsilon_{ilk}L_lx_k +\hbar^2 \varepsilon_{ilk}\varepsilon_{lmk}x_m = -2i\hbar \varepsilon_{ilk}L_lx_k$$

which is false. I used $$\varepsilon_{ilk}\varepsilon_{lkm}=-\varepsilon_{lik}\varepsilon_{lkm} = -\left(\delta_{ik}\delta_{km}-\delta_{im}\delta_{kk}\right) = 0$$ and $$\left[L_l, x_k\right] = i\hbar \varepsilon_{lkm}x_m$$.

Could you give me some hints? I'm starting to learn how to use Levi-Civita tensor

• First problem: after the "which leads me to" replace $j\rightarrow l$. – Sean E. Lake Mar 17 '20 at 14:49
• Next: What is $\epsilon_{ijk}\epsilon_{klm}$ in terms of Kronecker deltas? – Sean E. Lake Mar 17 '20 at 14:50
• @SeanE.Lake I edited the question. I have corrected the fact with the $j$ and the $l$. I continue having the same problem :( – Patrick Mar 17 '20 at 15:19

Found my error. $$\delta_{ik}\delta_{km}-\delta_{im}\delta_{kk} = (1-3)\delta_{im} \neq 0$$.