# How do you factor 125e^3-27f^3?

Jul 30, 2018

$\left(5e-3 f\right) \left(25 {e}^{2} + 15 e f + 9 {f}^{2}\right)$

#### Explanation:

$\text{this is a "color(blue)"difference of cubes}$

•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)

$125 {e}^{3} = {\left(5 e\right)}^{3} \Rightarrow a = 5 e$

$27 {f}^{3} = {\left(3 f\right)}^{3} \Rightarrow b = 3 f$

$= \left(5e-3 f\right) \left({\left(5 e\right)}^{2} + 5 e \cdot 3 f + {\left(3 f\right)}^{2}\right)$

$= \left(5e-3 f\right) \left(25 {e}^{2} + 15 e f + 9 {f}^{2}\right)$

Jul 30, 2018

$\left(5e-3 f\right) \left(25 {e}^{2} + 15 e f + 9 {f}^{2}\right)$

#### Explanation:

What we have is a difference of cubes, which factors as follows:

bar(ul|color(white)(2/2)(a^3-b^3)=(a-b)(a^2+ab+b^2)color(white)(2/2)|

We have the following:

$125 {e}^{3} - 27 {f}^{3}$, where $a = \sqrt[3]{125 {e}^{3}}$, and $b = \sqrt[3]{27 {f}^{3}}$.

Simplifying these, we get $a = 5 e$ and $b = 3 f$. Let's plug these into our difference of cubes expression to get

$\left(5e-3 f\right) \left(25 {e}^{2} + 15 e f + 9 {f}^{2}\right)$

Hope this helps!