# How do you factor x^3 - 81?

Oct 10, 2015

Express as a difference of cubes, then use the difference of cubes identity to find:

${x}^{3} - 81 = \left(x - {3}^{\frac{4}{3}}\right) \left({x}^{2} + {3}^{\frac{4}{3}} x + {3}^{\frac{8}{3}}\right)$

#### Explanation:

When $a \ge 0$ and $b , c \ne 0$

${a}^{b c} = {\left({a}^{b}\right)}^{c}$

So:

$81 = {3}^{4} = {3}^{\frac{4}{3} \cdot 3} = {\left({3}^{\frac{4}{3}}\right)}^{3}$

The difference of cubes identity is:

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

So:

${x}^{3} - 81 = {x}^{3} - {\left({3}^{\frac{4}{3}}\right)}^{3}$

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

$= \left(x - {3}^{\frac{4}{3}}\right) \left({x}^{2} + {3}^{\frac{4}{3}} x + {3}^{\frac{8}{3}}\right)$