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

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Answer 1 · 2015-07-02T17:49:46

#x^6-1 = (x^3-1)(x^3+1)#

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

Use difference of squares, difference of cubes and sum of cubes:

[1]: #a^2-b^2 = (a-b)(a+b)#

[2]: #a^3-b^3 = (a-b)(a^2+ab+b^2)#

[3]: #a^3+b^3 = (a+b)(a^2-ab+b^2)#

#x^6-1#

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

#=(x^3-1)(x^3+1)# by [1]

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

#=(x-1)(x^2+x+1)(x+1)(x^2-x+1)# by [2] and [3]