How do you factor $x^6-1$ ?

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Answer 1 · 2016-01-02T23:29:25

Use some standard identities to find:

$x^6-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$

Use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

the difference of cubes identity:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

and the sum of cubes identity:

$a^3+b^3=(a+b)(a^2-ab+b^2)$

as follows:

$x^6-1$

$=(x^3)^2-1^2$

$=(x^3-1)(x^3+1)$

$=(x^3-1^3)(x^3+1^3)$

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

That's as far as we can go with Real coefficients.

If we allow Complex coefficients then this factors further as:

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

where $omega = -1/2 + sqrt(3)/2 i$ is the primitive Complex cube root of $1$ .

How do you factor x^6-1x^6-1?