How do you factor #x^6-1#?
1 Answer
Jan 2, 2016
Use some standard identities to find:
#x^6-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)#
Explanation:
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
