# How do you factor 125x^3+144?

Dec 13, 2015

$125 {x}^{3} + 144 = \left(5 x + {144}^{\frac{1}{3}}\right) \left(25 {x}^{2} - 5 \left({144}^{\frac{1}{3}}\right) x + {144}^{\frac{2}{3}}\right)$

#### Explanation:

The sum of two cubes can be factored as
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

While unfortunately 144 is not a perfect cube, it still a cube of some number. Specifically it is the cube of ${144}^{\frac{1}{3}}$

Applying this, we have

$125 {x}^{3} + 144 = {\left(5 x\right)}^{3} + {\left({144}^{\frac{1}{3}}\right)}^{3}$

$= \left(5 x + {144}^{\frac{1}{3}}\right) \left(25 {x}^{2} - 5 \left({144}^{\frac{1}{3}}\right) x + {144}^{\frac{2}{3}}\right)$

With a sum of difference of cubes we always get a single real root, meaning we cannot factor any further. So our final factorization is

$125 {x}^{3} + 144 = \left(5 x + {144}^{\frac{1}{3}}\right) \left(25 {x}^{2} - 5 \left({144}^{\frac{1}{3}}\right) x + {144}^{\frac{2}{3}}\right)$