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)#