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

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Answer 1 · 2016-02-05T15:11:55

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

Start from the given:

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

by grouping method

first two terms, factor #x^6# and last two terms, factor the #-1#

that is

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

factor out the common binomial factor #(x^3-1)# so that

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

at this point, use " sum or difference of two cubes" forms
and difference of two squares

#a^3-b^3=(a-b)(a^2+ab+b^2)#
#a^3+b^3=(a+b)(a^2-ab+b^2)#
#a^2-b^2=(a-b)(a+b)#
so that

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

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

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

have a nice day ! from the Philippines...