Proof involving Poisson bracket

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Not being able to understand how each term has been simplified to get from the third step to the fourth step. So how did 1/2m become 1/m and {qj,plpl}pk become {qj,pl}plpk and how did k/4 become k/2 and {pk,(qlql)^2}qj become {pk,(qlql)}(qmqm)qj.

Any help would be much appreciated.

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The third to fourth line is obtained via the product rule again applied to both pieces: $$ \frac{1}{2m} \epsilon_{ijk} \{ q_j, p_l p_l \} p_k = \frac{1}{2m} \epsilon_{ijk} \{ q_j, p_l \} p_l p_k + \frac{1}{2m} \epsilon_{ijk} \{ q_j, p_l \} p_l p_k = \frac{1}{m}\epsilon_{ijk}\{q_j, p_l \} p_l p_k $$ Because both are the same statement it doesn't matter which $p_l$ is outside. The second piece: \begin{eqnarray*} \frac{k}{4} \epsilon_{ijk} \{ p_k, (q_l q_l)^2 \} q_j &=&\frac{k}{4} \epsilon_{ijk} \{ p_k, (q_l q_l)(q_m q_m) \} q_j \\ &=&\frac{k}{4}\left( \epsilon_{ijk} \{ p_k, (q_l q_l) \}(q_m q_m) q_j + \epsilon_{ijk} \{ p_k, (q_m q_m) \}(q_l q_l) q_j \right) \\ &=& \frac{k}{4}\left( \epsilon_{ijk} \{ p_k, (q_l q_l) \}(q_m q_m) q_j + \epsilon_{ijk} \{ p_k, (q_l q_l) \}(q_m q_m) q_j \right) \\ &=&\frac{k}{4}\left( 2\epsilon_{ijk} \{ p_k, (q_l q_l) \}(q_m q_m) q_j \right) = \frac{k}{2}\epsilon_{ijk} \{ p_k, (q_l q_l) \}(q_m q_m) q_j \end{eqnarray*} This last part follows from the product rule and then the fact that the repeated indices are not unique, just switch the l and m indices with each other in the second sum because they are dummy indices.