How do you factor completely #64-x^3#?

1 Answer
Sep 2, 2016

Answer:

#64-x^3 = (4-x)(16+4x+x^2)#

Explanation:

The difference of cubes identity can be written:

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

Note both #64=4^3# and #x^3# are perfect cubes.

So we find:

#64-x^3 = 4^3-x^3#

#color(white)(64-x^3) = (4-x)(4^2+4x+x^2)#

#color(white)(64-x^3) = (4-x)(16+4x+x^2)#