How do you factor #64x^3+8=0#?

1 Answer
mason m · Shwetank Mauria
Mar 7, 2016

#8(2x+1)(4x^2-2x+1)=0#

Explanation:

This is a sum of cubes, since both #64x^3=(4x)^3# and #8=(2)^3# are cubed terms.

Differences of cubes factor as follows:

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

In #64x^3+8#, we see that #a=4x# and #b=2#, so

#64x^3+8=0#

#(4x+2)((4x)^2-4x(2)+(2)^2)=0#

#(4x+2)(16x^2-8x+4)=0#

#8(2x+1)(4x^2-2x+1)=0#

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