How do you factor #r^3 - 1#?

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Answer 1 · 2015-12-20T00:32:58

Use the difference of cubes identity to find:

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

The difference of cubes identity can be written:

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

Use this with #a=r# and #b=1# as follows:

#r^3-1#

#=r^3-1^3#

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

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

If you allow Complex coefficients then this can be factored a little further:

#=(r-1)(r-omega)(r-omega^2)#

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