# How do you factor 2x^3-125?

Jun 24, 2017

$2 {x}^{3} - 125 = \left(\sqrt[3]{2} x - 5\right) \left(\sqrt[3]{4} {x}^{2} + 5 \sqrt[3]{2} x + 25\right)$

#### Explanation:

The difference of cubes identity can be written:

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

Note that $125 = {5}^{3}$ is a perfect cube (over the rationals), but $2 {x}^{3}$ is not.

We can treat it as a cube by using irrational coefficients, to find:

$2 {x}^{3} = {\left(\sqrt[3]{2} x\right)}^{3}$

and hence:

$2 {x}^{3} - 125 = {\left(\sqrt[3]{2} x\right)}^{3} - {5}^{3}$

$\textcolor{w h i t e}{2 {x}^{3} - 125} = \left(\sqrt[3]{2} x - 5\right) \left({\left(\sqrt[3]{2} x\right)}^{2} + \left(\sqrt[3]{2} x\right) \left(5\right) + {5}^{2}\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 125} = \left(\sqrt[3]{2} x - 5\right) \left(\sqrt[3]{4} {x}^{2} + 5 \sqrt[3]{2} x + 25\right)$

...noting that we have used $\sqrt[3]{a} \sqrt[3]{b} = \sqrt[3]{a b}$ and hence:

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