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Given that B = CAC-1 & CA5C-1 = Bm, m\(\in \)N

B2 = (CAC-1) (CAC-1) = CA(C-1C) AC-1 = CAIAC-1

 = C.A.A.C-1 = CA2C-1

B4 = (CA2C-1) (CA2C-1) = CA2(C-1C)A2C-1

B5 = (CA4C-1) (CAC-1) = CA4(C-1C)AC-1

 = CA5C-1

Hence, m = 5

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