# How do you factor 54x^3-16y^3?

Mar 26, 2016

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

#### Explanation:

There is a common factor of 2 , thus

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

Now $27 {x}^{3} - 8 {y}^{3} \text{ is a difference of cubes }$

which factors as follows:

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

For $27 {x}^{3} , a = 3 x \text{ and for } 8 {y}^{3} , b = 2 y$

$\Rightarrow 27 {x}^{3} - 8 {y}^{3} = \left(3 x - 2 y\right) \left(9 {x}^{2} + 6 x y + 4 {y}^{2}\right)$

$\Rightarrow 54 {x}^{3} - 16 {y}^{3} = 2 \left(3 x - 2 y\right) \left(9 {x}^{2} + 6 x y + 4 {y}^{2}\right)$