Question

show that [A,[B,C]]+[B,[C,A]+[C,[A,B]]=0

Answer 1

Using the definition of commutator of two operators, $[A, B] = A B - B A$, let us open the nested commutators:

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

$= A (BC - CB) - (BC - CB)A + B(CA - AC) - (CA - AC) B + C (AB - BA) - (AB - BA) C$ Opening the brackets, and regrouping the terms, obtain:

$(ABC - ABC) + (- ACB + ACB) + (-BCA + BCA) + (CBA - CBA) + (-BAC + BAC) + (-CAB + CAB) = 0$