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

Question

6554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-07T12:24:12

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

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

$64x^3+8=0$

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

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

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

How do you factor 64x^3+8=064x^3+8=0?