Question

How do you factor `x^9 - x^6 - x^3 + 1`?

Answer 1

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

Explanation:

`x^9-x^6-x^3-1`
`x^6(x^3-1)-1(x^3-1)`
`(x^3-1)(x^6-1)`
`(x^3-1)[(x^3)^2-(1^2)]`
`(x^3-1)(x^3-1)(x^3+1)`
#(x^3-1)^2(x^3+1) #

Answer 2

`(x+1)(x-1)^2(x^2-x+1)(x^2+x+1)^2`

Explanation:

Let's revise our power rules ( they will come in handy later ):

1. Difference of squares rule: `x^2-y^2=(x-y)(x+y)`
2. Difference of cubes rule: `x^3-y^3=(x-y)(x^2+xy+y^2)`
3. Sum of cubes rule: `x^3+y^3=(x+y)(x^2-xy+y^2)`

`x^9 - x^6 - x^3 + 1`

Factorise `x^6` from the first two terms,

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

Factorise `(x^3-1)`,

`(x^6-1)(x^3-1)`

Apply difference of squares rule to `(x^6-1)`,

`color(red)((x^3+1))color(blue)((x^3-1))color(green)((x^3-1))`

Apply sum of cubes rule to `(x^3+1)`,

`color(red)((x+1)(x^2-x+1))color(blue)((x^3-1))color(green)((x^3-1))`

Apply difference of cubes rule to `(x^3-1)`,

`color(red)((x+1)(x^2-x+1))color(blue)((x-1)(x^2+x+1))color(green)((x-1)(x^2+x+1))`

Simplify like polynomials,

`(x+1)(x-1)^2(x^2-x+1)(x^2+x+1)^2`

There you go.

P.S. I did not include the sum of squares rule because that rule dwells into imaginary numbers but if you are keen to know, here it is: `x^2+y^2=(x+yi)(x-yi)`