Answer:
Three linear operators A,B, and C will satisfy the condition [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].
Explanation:
According to the question we have to prove.
[tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex]
Now taking Left hand side of the equation and solve.
[tex][[A, B],C] + [[B,C), A] + [[C, A], B][/tex]
Now use commutator property on it as,
[tex]=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0[/tex]
Therefore, it is proved that [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].
