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Rewrite x^{8}-1 as \left(x^{4}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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Using the formula we get $$ x^8 -1$$ $$ = (x^2 +1 + \sqrt {2}x) (x^2 +1- \sqrt {2}x)(x^2 +1)(x+1)(x-1) $$

This problem is called a "Difference of two Squares" since x^8 is actually a perfect square x^4*x^4: x^8-1 (x^4-1)(x^4+1) The SUM of two Squares does NOT factor like the Difference of Squares factors, so leave it alone. However, notice that the first factor x^4-1 DOES factor again: (x^2-1)(x^2+1)(x^4+1)

We have to break the term x 8 − 1 in the product of its factors such that when multiplied, they give us the value x 8 − 1. Let us write power 8 as product of 4 and 2. ∴ The factors of x 8 − 1 are (x − 1); (x + 1); (x 2 + 1); (x 4 + 1).