# How do you factor (3x-4)^3 + 27?

Apr 10, 2015

This is a sum of two cubes. Memorize the rule:

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

In this example, $a = \textcolor{red}{\left(3 x - 4\right)}$ and $b = 3$, so

$\textcolor{red}{{\left(3 x - 4\right)}^{3}} + {3}^{3} = \left(\textcolor{red}{\left(3 x - 4\right)} + 3\right) \left(\textcolor{red}{{\left(3 x - 4\right)}^{2}} - \textcolor{red}{\left(3 x - 4\right)} 3 + {3}^{2}\right)$

This answer can be simplified to get:

$\left(3 x - 1\right) \left({x}^{2} - 24 x + 16 - 9 x + 12 + 9\right) = \left(3 x - 1\right) \left({x}^{2} - 33 x + 37\right)$