Question

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

Answer 1

You can use the difference of cubes identity twice to partially factor this as:

`1-x^9 = (1-x)(1+x+x^2)(1+x^3+x^6)`

or more thoroughly to find:

`1-x^9 = (1-x)(1+x+x^2)(1-2 cos((2pi)/9)x + x^2)(1-2 cos((4pi)/9)x+x^2)(1-2 cos((8pi)/9)x+x^2)`

Explanation:

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`