Question 20703
x^8 - 1
=(x^2)^3 - 1^3     
applying the formula for a^3 - b^3

=(x^2 - 1)((x^2)^2 + 1 + x^2)
=(x^2 - 1^2)((x^2)^2 + 1 + x^2)
=(x-1)(x+1)(x^2)^2 + 1 + x^2)

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