How do you factor completely #x^3 - 1#?

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Answer 1 · 2016-01-03T17:04:28

Using the identity #a^3-b^3=(a-b)*(a^2+ab+b^2)# we have that

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

Answer 2 · 2016-01-03T17:07:47

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

This is a difference of cubes.

Differences of cubes can be factored as follows:

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

In your situation, #a=x# and #b=1#, since #a^3=x^3# and #b^3=1#.

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

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