Question

1000 points-------->
factorise
x^8-1

Answer 1

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Answer 2

$x^8 - 1 \\ \\ = (x^4)^2 - 1^2 \\ \\ = (x^4 + 1)(x^4 - 1) \\ \\ = (x^4 + 1)((x^2)^2 - 1^2) \\ \\ = (x^4 + 1)(x^2 + 1)(x^2 - 1) \\ \\ = (x^4 + 1)(x^2 + 1)(x^2 - 1^2) \\ \\ = (x^4 + 1)(x^2 + 1)(x + 1)(x - 1)$

$We\ know\ that, \\ \\ i = \sqrt{-1} \\ \\ i^2 = -1$

$\\ \\ \\ \therefore\ (x^4 + 1)(x^2 + 1)(x + 1)(x - 1) \\ \\ = (x^4 - (-1))(x^2 - (-1))(x + 1)(x - 1) \\ \\ = ((x^2)^2 - i^2)(x^2 - i^2)(x + 1)(x - 1) \\ \\ = (x^2 + i)(x^2 - i)(x + i)(x - i)(x + 1)(x - 1) \\ \\ \\ \therefore\ \underline{\underline{x^8 - 1 = (x^2 + i)(x^2 - i)(x + i)(x - i)(x + 1)(x - 1)}}$

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