If A and B are different matrices satisfying A3 = B3 and A2B = B2A, then
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A3 = B3 ....(1)
and A2B = B2A ....(2)
(1) – (2) gives, A3 – A2B = B3 – B2A
A2(A – B) = – B2(A – B) ⇒ (A2 + B2)(A – B) = 0
det (A2 + B2)(A – B) = 0
det (A2 + B2) · det (A – B) = 0