# How do you factor the expression 81x^6+24x^3y^3?

Apr 10, 2018

$81 {x}^{6} + 24 {x}^{3} {y}^{3} = 3 {x}^{3} \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

#### Explanation:

Given:

$81 {x}^{6} + 24 {x}^{3} {y}^{3}$

Note that both factors are divisible by $3 {x}^{3}$. So we can separate out that factor first to find:

$81 {x}^{6} + 24 {x}^{3} {y}^{3} = 3 {x}^{3} \left(27 {x}^{3} + 8 {y}^{3}\right)$

Next note that both $27 {x}^{3} = {\left(3 x\right)}^{3}$ and $8 {y}^{3} = {\left(2 y\right)}^{3}$ are perfect cubes. So we can use the sum of cubes identity:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

with $A = 3 x$ and $B = 2 y$ to find:

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

$\textcolor{w h i t e}{27 {x}^{3} + 8 {y}^{3}} = \left(3 x + 2 y\right) \left({\left(3 x\right)}^{2} - \left(3 x\right) \left(2 y\right) + {\left(2 y\right)}^{2}\right)$

$\textcolor{w h i t e}{27 {x}^{3} + 8 {y}^{3}} = \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$

Putting it all together:

$81 {x}^{6} + 24 {x}^{3} {y}^{3} = 3 {x}^{3} \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$