# How do you factor 64x^3 + 27y^3?

Apr 30, 2016

$\left(4 x + 3 y\right) \left(16 {x}^{2} - 12 x y + 9 {y}^{2}\right)$

#### Explanation:

This expression is a $\textcolor{b l u e}{\text{ sum of cubes "" and is factorised as follows }}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

now $64 {x}^{3} = {\left(4 x\right)}^{3} \text{ and } 27 {y}^{3} = {\left(3 y\right)}^{3}$

For factorising purposes a = 4x and b = 3y

$\Rightarrow 64 {x}^{3} + 27 {y}^{3} = \left(4 x + 3 y\right) \left(16 {x}^{2} - 12 x y + 9 {y}^{2}\right)$