# How do you factor z^3 - 4?

Apr 26, 2016

${z}^{3} - 4 = \left(z - \sqrt[3]{4}\right) \left({z}^{2} + \sqrt[3]{4} z + 2 \sqrt[3]{2}\right)$

#### Explanation:

This can be factored as a difference of cubes.

The difference of cubes identity can be written:

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

We can use this with $a = z$ and $b = \sqrt[3]{4}$ as follows:

${z}^{3} - 4$

$= {z}^{3} - {\left(\sqrt[3]{4}\right)}^{3}$

$= \left(z - \sqrt[3]{4}\right) \left({z}^{2} + \sqrt[3]{4} z + \sqrt[3]{16}\right)$

$= \left(z - \sqrt[3]{4}\right) \left({z}^{2} + \sqrt[3]{4} z + 2 \sqrt[3]{2}\right)$