Questions
If then
detailed solution
Correct option is D
q=pn for some p∈Nzq−1=znp−1=zn−1z(p−1)n+⋯+zn+1Every root of zn−1 is a root of zq−1 and every root of z(p−1)n+⋯+zn+1=0 is also a root of zq−1 Hence. A⊂B∣and A∩B=ASimilar Questions
If and then
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