The Poisson bracket possesses the symplectic structure \([A,B] = AB - BA.\)

(Note that this algebraic structure is not commutative)

Apply this rule to the brackets:

\[[A, [B,C]],\] \[[B, [C,A]],\] \[[C, [A,B]].\]

Assemble the three brackets to prove that \[[A,[B,C]] + [B, [C,A]] + [C, [A,B]] = 0.\]

**Solution**

Following this result, we apply the same rules to the following brackets:

\[\left[ A,\left[ B,C \right] \right] = \left[ A,\left[ BC-CB \right] \right] = A(BC-CB) - (BC-CB)A=ABC-ACB-BCA+CBA\]

\[\left[ B,\left[ C,A \right] \right] = \left[ B,\left[ CA-AC \right] \right] = B(CA-AC) - (CA-AC)B=BCA-BAC-CAB+ACB \]

\[\left[ C,\left[ A,B \right] \right] =\left[ C,\left[ AB-BA \right] \right] = C(AB-BA) - (AB-BA)C=CAB-CBA-ABC+BAC \]

We note that the order of \(A,B,C\) is significant for the bracket operator. Adding the three brackets will yield 0.

Check out my other notes at Proof, Disproof, and Derivation

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