# How do you factor 27x^3-8y^3?

Apr 1, 2018

The factored polynomial is $\left(3 x - 2 y\right) \left(9 {x}^{2} + 6 x y + 4 {y}^{2}\right)$.

#### Explanation:

You can use the difference of cubes factoring pattern:

${\textcolor{red}{a}}^{3} - {\textcolor{b l u e}{b}}^{3} \quad = \quad \left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right) \left({\textcolor{red}{a}}^{2} + \textcolor{red}{a} \textcolor{b l u e}{b} + {\textcolor{b l u e}{b}}^{2}\right)$

Right now, our terms are ${a}^{3} = 27 {x}^{3}$ and ${b}^{3} = 8 {y}^{3}$. Take the cube root of each to find that $a = 3 x$ and $b = 2 y$. Now plug them into the difference of cubes:

$\textcolor{w h i t e}{=} 27 {x}^{3} - 8 {y}^{3}$

$= {\left(\textcolor{red}{3 x}\right)}^{3} - {\left(\textcolor{b l u e}{2 y}\right)}^{3}$

$= \left(\textcolor{red}{3 x} - \textcolor{b l u e}{2 y}\right) \left({\left(\textcolor{red}{3 x}\right)}^{2} + \left(\textcolor{red}{3 x}\right) \left(\textcolor{b l u e}{2 y}\right) + {\left(\textcolor{b l u e}{2 y}\right)}^{2}\right)$

$= \left(\textcolor{red}{3 x} - \textcolor{b l u e}{2 y}\right) \left({\textcolor{red}{3}}^{2} {\textcolor{red}{x}}^{2} + \textcolor{p u r p \le}{6} \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{2}}^{2} {\textcolor{b l u e}{y}}^{2}\right)$

$= \left(\textcolor{red}{3 x} - \textcolor{b l u e}{2 y}\right) \left({\textcolor{red}{9 x}}^{2} + \textcolor{p u r p \le}{6} \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{4 y}}^{2}\right)$

$= \left(3 x - 2 y\right) \left(9 {x}^{2} + 6 x y + 4 {y}^{2}\right)$

This is factored all the way. Hope this helped!