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

#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) #

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)#