The HCF of x8 – 1 and x4 + 2x3 – 2x

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The HCF of x⁸ – 1 and x⁴ + 2x³ – 2x – 1 is :

1. x² + 1
2. x² - 1
3. x + 1
4. x - 1

Correct Option: B

x⁸ – 1 = (x⁴)² - 1²
= (x⁴ + 1)(x⁴ - 1)
= (x⁴ + 1)(x² - 1) (x² + 1)
= (x⁴ + 1)(x² + 1)(x + 1)(x - 1)
[ ∴ a² - b² = (a + b)(a - b) ]
x⁴ + 2x³ – 2x – 1 = (x⁴ - 1) + 2x³ – 2x
= (x² - 1) (x² + 1) + 2x(x² - 1)
= (x² + 1 + 2x)(x² - 1)
= (x + 1)²(x + 1)(x - 1)
∴ H.C.F = (x + 1) (x – 1) = (x² - 1)