Answer
See below
Work Step by Step
If $A, B$, and $C$ are $n\times n$ matrices, then:
$[A,[B,C]]\\=[A,,BC-CB]\\=A(BC-CB)-A(BC-CB)\\=ABC-ACB-BCA+CBA$
Obtain: $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]\\=[A,BC-CB]+[B,CA-AC]+[C,AB-BA]\\=A(BC-CB)-(BC-CB)A+B(CA-AC)-B(CA-AC)+C(AB-BA)-C(AB-BA)\\=ABC-ACB-BCA+CBA+BCA-BAC-CAB+ACB+CAB-CBA-ABC+BAC\\=0$
