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

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

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Answer 1 · 2016-03-07T12:24:12

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

Explanation:

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

Differences of cubes factor as follows:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$

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

$64 x^{3} + 8 = 0$ $64x^3+8=0$

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

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

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

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

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