Question
Factorise:
$x^8- 1$
$x^8- 1$
✓
Answer
$x^8- 1 = (x^4)^2 - (1)^2$
$= (x^4 - 1)(x^4 + 1)$
$= [(x^2)^2 - (1)^2)(x^4 + 1)$
$= (x^2 - 1)(x^2 + 1)(x^4 + 1)$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2 - 2x^2$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2) - 2x^2$
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\\\ \ \ \ \ \Big[\big(\text{x}^2+1) -\big(\sqrt{2}\text{x}\big)^2\Big] $
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\big(\text{x}^2+1-\sqrt{2}\text{x}\big)\\\ \ \ \ \big(\text{x}^2+1+\sqrt{2}\text{x}\big) $
$= (x^4 - 1)(x^4 + 1)$
$= [(x^2)^2 - (1)^2)(x^4 + 1)$
$= (x^2 - 1)(x^2 + 1)(x^4 + 1)$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2 - 2x^2$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2) - 2x^2$
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\\\ \ \ \ \ \Big[\big(\text{x}^2+1) -\big(\sqrt{2}\text{x}\big)^2\Big] $
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\big(\text{x}^2+1-\sqrt{2}\text{x}\big)\\\ \ \ \ \big(\text{x}^2+1+\sqrt{2}\text{x}\big) $
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