# How do you factor 8x^3 - 1?

##### 1 Answer
Apr 9, 2015

Remember this formula for factorizing difference of 2 cubes:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

In $8 {x}^{3} - 1$,

${a}^{3} = 8 {x}^{3}$
${b}^{3} = 1$

$a = \sqrt[3]{8 {x}^{3}}$ = $2 x$
$b = \sqrt[3]{1}$= $1$

Substitute $a = 2 x$ , $b = 1$ into the formula of $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

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

$\left(2 x - 1\right)$(4x^2 + $2 x$+ 1) is the factorized form of $8 {x}^{3} - 1$