How do I factor $8 x^{3} - 1$ $8x^3-1$ completely?

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How do I factor $8 x^{3} - 1$ $8x^3-1$ completely?

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Answer 1 · 2015-07-26T18:11:13

Factor 8x^3 - 1

Explanation:

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

$8 x^{3} - 1 = (2 x - 1) (4 x^{2} + 2 x + 1)$ $8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1)$

Answer 2 · 2015-07-26T18:14:28

You use the formula for the difference of perfect cubes.

Explanation:

All you really have to do in order to factor this expression completely is use the formula for the difference of two perfect cubes

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

You can rewrite the original expression as

$8 x^{3} - 1 = {(2 x)}^{3} - 1^{3}$ $8x^3 - 1 = (2x)^3 - 1^3$

This will get you

${(2 x)}^{3} - 1^{3} = (2 x - 1) [{(2 x)}^{2} + 2 x \cdot 1 + 1^{2}]$ $(2x)^3 - 1^3 = (2x - 1)[(2x)^2 + 2x * 1 + 1^2]$

${(2 x)}^{3} - 1^{3} = (2 x - 1) \cdot (4 x^{2} + 2 x + 1)$ $(2x)^3 - 1^3 = (2x - 1) * (4x^2 + 2x + 1)$

The quadratic $4 x^{2} + 2 x + 1$ $4x^2 + 2x + 1$ cannot be factored further without using complex numbers, which I'm not sure you're supposed to use.

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How do I factor $8 x^{3} - 1$ $8x^3-1$ completely?

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