Question

How do you factor completely `x^9-1`?

Answer 1

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

Explanation:

To factorize `x^9-1`, we can use the identity

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

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

factorizing `x^3-1` further

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

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

Answer 2

The separate factors are `(x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8`. The first factor `(x-1)` is the only real factor.

Explanation:

The roots of this equation `x^9-1=0` are the 9 values of `x=1^(1/9)`.
They are`(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8`.
k=0 gives the only real root 1.

So, the factors of `x^0-1` are `(x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8`.