Question

How do you factor `x^9 - 27`?

Answer 1

You can use the difference of cubes identity to help factor this, but to find all the factors with Real coefficients requires a little more...

Explanation:

You can partially factor `x^9-27` by first recognising that it is a difference of cubes.

The difference of cubes identity can be written:

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

In our case we can put `a=x^3` and `b=3` to find:

`x^9-27 = (x^3)^3-3^3 = (x^3-3)((x^3)^2+(x^3)(3)+3^2)`

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

The remaining cubic factor can itself be treated as a difference of cubes:

`x^3-3 = x^2-(root(3)(3))^3 = (x-root(3)(3))(x^2+root(3)(3)x+root(3)(3)^2)`

The remaining sextic factor can be factored as the product of three quadratics:

`x^6+3x^3+9 =`

`(x^2-2cos((2pi)/9)root(3)(3)x+root(3)(3)^2) *`

`(x^2-2cos((4pi)/9)root(3)(3)x+root(3)(3)^2) *`

`(x^2-2cos((8pi)/9)root(3)(3)x+root(3)(3)^2)`

To see why requires some Complex arithmetic.

The primitive Complex `9`th root of `1` is:

`alpha = cos((2pi)/9) + i sin((2pi)/9)`

Then all the `9`th roots of `1` are `alpha, alpha^2,..., alpha^9 = 1`

Apart from `alpha^9 = 1`, these root occur in Complex conjugate pairs, which we can combine to find quadratic factors with Real coefficients. The roots `alpha^3`, `alpha^6` and `alpha^9=1` are all cube roots of `1` relating to the factors of `x^3-3`.

The other `6` roots combine in pairs to give the three other quadratic factors of the sextic:

`(x-alpha root(3)(3))(x-alpha^8 root(3)(3)) = x^2-2 cos((2pi)/9)root(3)(3)x + root(3)(3)^2`

`(x-alpha^2 root(3)(3))(x-alpha^7 root(3)(3)) = x^2-2 cos((4pi)/9)root(3)(3)x + root(3)(3)^2`

`(x-alpha^4 root(3)(3))(x-alpha^5 root(3)(3)) = x^2-2 cos((8pi)/9)root(3)(3)x + root(3)(3)^2`