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

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Answer 1 · 2016-09-02T21:11:22

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

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:

$64-x^3 = 4^3-x^3$

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

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

How do you factor completely 64-x^364-x^3?