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SOLVED: (b) Show that LL=0 Hint: The following commutator identities are helpful: [B,A]=-[A,B] [A,A]=0 [A,B+C]=[A,B]+[A,C] [A+B,C]=[A,C]+[B,C] [A,BC]=[A,B]C+B[A,C] [AB,C]=[A,C]B+A[B,C] [AB,CD]=[A,C]BD+A[B,C]D+C[A,D]B+AC[B,D]