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

Oct 4, 2016

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

#### Explanation:

This is the difference of cubes and has a set pattern.

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

make 2 brackets:
The first is simply the cube root of each term

The second bracket is formed from the first in 5 steps:

$1. \rightarrow$ square the first term.
$2. \rightarrow$ change the sign
$3. \rightarrow$ multiply the two terms
$4. \rightarrow$ PLUS
$5. \rightarrow$square the second term

Oct 4, 2016

$27 \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$

#### Explanation:

The first step here is to take out a $\textcolor{b l u e}{\text{common factor}}$ of 27.

$\Rightarrow 27 \left({x}^{3} - {y}^{3}\right) \ldots \ldots . . \left(A\right)$

Now ${x}^{3} - {y}^{3}$ is a $\textcolor{b l u e}{\text{difference of cubes}}$ and in general factorises as.

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

here a = x and b = y

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

Substituting this result into (A) gives the factorised form.

$\Rightarrow 27 {x}^{3} - 27 {y}^{3} = 27 \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$