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This

$$[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0$$

This proof is not wanted, since there is an attachment of some variables.

How can you prove the equation?

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closed as off-topic by Waffle's Crazy Peanut, Emilio Pisanty, Manishearth Sep 18 '13 at 18:39

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    $\begingroup$ Is $[\cdot,\cdot]$ supposed to be the Poisson bracket as it is in the proof you referenced or do you mean the quantum mechanical commutator? $\endgroup$ – Jonas Greitemann Sep 18 '13 at 12:10
  • $\begingroup$ In general you can't. But if [,] is the commutator, why dont just just expand the Jacobi formula? $\endgroup$ – jinawee Sep 18 '13 at 12:13
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    $\begingroup$ I'd consider this a math question. $\endgroup$ – Nikolaj-K Sep 18 '13 at 12:25
  • $\begingroup$ Comments to the question (v1): Echoing Jonas's comment, is this a question about Poisson brackets or operator commutators? The proof is very different depending on which. And what is 'generalized' about the Jacobi identity? $\endgroup$ – Qmechanic Sep 18 '13 at 15:58
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You can use the definition of the QM commutator $[X,Y]=XY-YX$, then expand all the commutators and simplify.

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