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

Mar 29, 2018

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

#### Explanation:

${x}^{3} - 8 {y}^{3}$
$\Rightarrow {\left(x\right)}^{3} - {\left(2 y\right)}^{3}$
$\Rightarrow {\left(x - 2 y\right)}^{3} + 3. x .2 y \left(x - 2 y\right)$
$\Rightarrow \left(x - 2 y\right) \left[{\left(x - 2 y\right)}^{2} + 6 x y\right]$
$\Rightarrow \left(x - 2 y\right) \left[{x}^{2} - 4 x y + 4 {y}^{2} + 6 x y\right]$
$\Rightarrow \left(x - 2 y\right) \left({x}^{2} + 2 x y + 4 {y}^{2}\right)$

Mar 29, 2018

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

#### Explanation:

${x}^{3} - 8 {y}^{3} \leftarrow \textcolor{b l u e}{\text{is a difference of cubes}}$

•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)

$\text{here "a=x" and } b = 2 y \to {\left(2 y\right)}^{3} = 8 {y}^{3}$

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

Mar 29, 2018

x^3-8y^3=color(blue)((x-2y)(x^2+2xy+4y^2)

#### Explanation:

Factor:

${x}^{3} - 8 {y}^{3}$

Apply the difference of cubes:

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

where:

$a = x$, and $b = 2 y$

Plug in the known values.

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

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

Apply multiplicative distributive property: ${\left(a b\right)}^{m} = {a}^{m} {b}^{m}$

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

Simplify.

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