How do you factor #64x^6 -1#?

1 Answer
Jan 19, 2016

Answer:

Use some standard identities to find:

#64x^6-1=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1)#

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

We find:

#64x^6-1#

#=(8x^3)^2-1^2#

#=(8x^3-1)(8x^3+1)#

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

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

#=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1)#

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this factors further as:

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

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