# How do I factor 8x^3-1 completely?

Jul 26, 2015

Factor 8x^3 - 1

#### Explanation:

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

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

Jul 26, 2015

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

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

You can rewrite the original expression as

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

This will get you

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

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

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