# How to factor 16x^9 + 54y^3 using difference of cubes?

Oct 10, 2017

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

#### Explanation:

Here we have sum of cubes and not difference of cubes. Hence, we have to use ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$. As such

$16 {x}^{9} + 54 {y}^{3}$

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

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

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

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