What is the HCF of 8 (N5 - N3 + N ) and 28 (N6 + 1)?

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What is the HCF of 8 (N⁵ - N³ + N ) and 28 (N⁶ + 1)?

1. 4 (N⁴ - N² + 1)
2. N³ - N + 4N²
3. N³ - N + 3N²
4. None of these

Correct Option: A

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)