# How do you factor (p-5)^3+125?

Jan 23, 2016

$p \left({p}^{2} - 15 p + 75\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Here, we have

${\left(p - 5\right)}^{3} + 125$

$= {\left(p - 5\right)}^{3} + {5}^{3}$

$= \left(\left(p - 5\right) + 5\right) \left({\left(p - 5\right)}^{2} - \left(p - 5\right) \left(5\right) + {5}^{2}\right)$

$= p \left({p}^{2} - 10 p + 25 - 5 p + 25 + 25\right)$

$= p \left({p}^{2} - 15 p + 75\right)$

The internal quadratic ${p}^{2} - 15 p + 75$ cannot be factored without using complex numbers.