# How do you factor 8a^3-27?

Feb 2, 2016

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

#### Explanation:

1) Decide factoring method

In the equation both $8$ and $27$ are cubes so we can use the Difference of Cubes method of factoring

2) Solve for variables

The formula for the Difference of Cubes method is:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

First we find $a$:
${a}^{3} = 8 {a}^{3}$
$\sqrt[3]{{a}^{3}} = \sqrt[3]{8 {a}^{3}}$
$a = 2 a$

Then we find $b$:
${b}^{3} = 27$
$\sqrt[3]{{b}^{3}} = \sqrt[3]{27}$
$b = 3$

3) Fill in formula

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

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

4) Simplify

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