How do you factor #64r^3-p^3#?

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Answer 1 · 2016-04-23T22:16:56

#64r^3-p^3=(4r-p)(16r^2+4rp+p^2)#

The difference of cubes identity can be written:

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

We can use this with #a=4r# and #b=p# as follows:

#64r^3-p^3#

#=(4r)^3-p^3#

#=(4r-p)((4r)^2+(4r)p+p^2)#

#=(4r-p)(16r^2+4rp+p^2)#

This cannot be factored further with Real coefficients.

If you allow Complex coeffients then it can be factored a little further:

#= (4r-p)(4r-omega p)(4r-omega^2 p)#

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