What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?

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Accepted Answer

Let p(N) = 8 (N5 - N3 + N) = 4 x 2 x N (N4 - N2 + 1) and q(N) = 28 (N6 + 1) = 7 x 4 [( N2)3 + (1)3] = 4 x 7 (N2 + 1) ( N4 - N2 + 1) ∴ HCF of p(N) and q (N) = 4 (N4 - N2 + 1)

What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?

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