# How do you factor 32x^3 - 4?

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

#### Explanation:

the given : $32 {x}^{3} - 4$

common monomial factor $= 4$
factor out $4$ first

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

use the "difference of two cubes " form

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

for $8 {x}^{3} - 1$

so that

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

$32 {x}^{3} - 4 = 4 \cdot \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$