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

2 Answers

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)#

Jan 3, 2016

Answer:

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

Explanation:

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)#