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

1 Answer
Jan 2, 2016

Answer:

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