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

1 Answer
Dec 20, 2015

Answer:

Use the sum of cubes identity to find:

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

Explanation:

The sum of cubes identity may be written:

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

In our example, we have #a=1# and #b=x# as follows:

#1+x^3#

#=1^3+x^3#

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

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

The remaining quadratic factor #(1-x+x^2)# cannot be factored into simpler factors with Real coefficients, but if you want a complete factorisation then you can do it with Complex coefficients:

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

or if you prefer:

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

where #omega = -1/2+sqrt(3)/2 i# is the primitive Complex cube root of #1#.