# How do you factor (a+2b)^3 - (a-2b)^3?

Apr 15, 2015

Remembering that a difference of cubes can be factored:

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

then:

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

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

$= \left(a + 2 b - a + 2 b\right) \left({a}^{2} + 4 a b + 4 {b}^{2} + {a}^{2} - 4 {b}^{2} + {a}^{2} - 4 a b + 4 {b}^{2}\right] =$

$= 4 b \left(3 {a}^{2} + 4 {b}^{2}\right)$.

But, in this case, there is a faster and easier way:

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

$= {a}^{3} + 6 {a}^{2} b + 12 a {b}^{2} + 8 {b}^{3} - {a}^{3} + 6 {a}^{2} b - 12 a {b}^{2} + 8 {b}^{3} =$

$= 12 {a}^{2} b + 16 {b}^{3} = 4 b \left(3 {a}^{2} + 4 {b}^{2}\right)$.