# How do you factor 27x^3+8y^3?

May 15, 2018

$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

#### Explanation:

$27 {x}^{3} + 8 {y}^{3}$

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

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

=$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

May 15, 2018

$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

#### Explanation:

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

May 15, 2018

$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

#### Explanation:

$\text{this is a "color(blue)"sum of cubes}$

$\text{which factors in general as}$

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

$27 {x}^{3} = {\left(3 x\right)}^{3} \Rightarrow a = 3 x$

$8 {y}^{3} = {\left(2 y\right)}^{3} \Rightarrow b = 2 y$

$\Rightarrow 27 {x}^{3} + 8 {y}^{3} = \left(3 x + 2 y\right) \left({\left(3 x\right)}^{2} - \left(3 x \times 2 y\right) + {\left(2 y\right)}^{2}\right)$

$\textcolor{w h i t e}{\times \times \times \times \times} = \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$