How do you factor #(1-x^9)#?

The difference of cubes identity is:

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

We can use this identity to partially factor #1-x^9# as follows:

#1-x^9 = 1^3-(x^3)^3 = (1-x^3)(1+x^3+(x^3)^2))#

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

#= (1-x)(1+x+x^2)(1+x^3+x^6)#

It is possible to factor #1+x^3+x^6# further into quadratics with Real coefficients, expressible in terms of #cos#:

#1+x^3+x^6 = (1-2 cos((2pi)/9)x + x^2)(1-2 cos((4pi)/9)x+x^2)(1-2 cos((8pi)/9)x+x^2)#

To understand how this works, note that De Moivre's Theorem gives us:

#(cos x + i sin x)^n = cos(nx) + i sin(nx)#

Putting #n = 9# and #x = (2k pi)/9#, where #k in ZZ# we find:

#(cos ((2k pi)/9) + i sin ((2k pi)/9))^9 = 1#

This gives us #9# distinct #9#th roots of #1#.

Every third one of these roots is a cube root of #1#, so a zero of #1-x^3 = (1-x)(1+x+x^2)#. The others are zeros of the remaining factor #(1+x^3+x^6)# These #6# remaining zeros occur in Complex conjugate pairs which combine to give factors with Real coefficients.

For example:

#(x-cos((2pi)/9)-i sin((2pi)/9))(x-cos((16pi)/9)-i sin((16pi)/9))#

#=(x-cos((2pi)/9)-i sin((2pi)/9))(x-cos((2pi)/9)+i sin((2pi)/9))#

#=x^2-2cos((2pi)/9)x+(cos^2((2pi)/9) + sin^2((2pi)/9))#

#=x^2-2cos((2pi)/9)x+1#

Similarly for #(4pi)/9# and #(8pi)/9#