# How do you factor x^3+x^2-125p^3-25p^2?

Sep 23, 2016

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

#### Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

So we find:

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

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

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

This has no simpler factors.