How do you factor $64x^3-8$ ?

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Answer 1 · 2017-02-20T02:54:52

$8(2x-1)(4x^2+2x+1)$

Factor $64x^3-8$

First factor out the GCF $8$ .

$8(8x^3-1)$

Rewrite the expression as a difference of cubes, where $a=2x$ and $b=1$ .

$8((2x)^3-1^3)$

Use the difference of cubes to answer this question:
$a^3-b^3=(a-b)(a^2+ab+b^2)$

$8((2x)^3-1^3)=8(2x-1)((2x^2+(2x*1)+1^2)=$

$8(2x-1)(4x^2+2x+1)=$

$8(2x-1)(4x^2+2x+1)$

How do you factor 64x^3-864x^3-8?