How do you factor #64x^6 -1#?
1 Answer
Jan 19, 2016
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
