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 $1$ .

How do you factor 64r^3-p^364r^3-p^3?