# How do you factor x^6-27?

##### 2 Answers
Aug 11, 2016

${x}^{6} - 27 = \left({x}^{2} - 3\right) \left({x}^{4} + 3 {x}^{2} + 9\right)$

#### Explanation:

The difference of two perfect cubes has a pattern which just has to be applied once you recognise that these terms are both cubes.

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

${x}^{6} - 27 = \left({x}^{2} - 3\right) \left({x}^{4} + 3 {x}^{2} + 9\right)$

First bracket: $\left(\sqrt[3]{x} \text{ same sign } \sqrt[3]{y}\right) \Rightarrow \left(x - y\right)$

Form the second bracket from the first.

" square the first term " " change sign" " product of the terms " "PLUS " "square the second term ")

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

Aug 11, 2016

${x}^{6} - 27 = \textcolor{g r e e n}{\left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left({x}^{4} + 3 {x}^{2} + 9\right)}$

#### Explanation:

Temporarily replacing ${x}^{2}$ with $k$, we have
$\textcolor{w h i t e}{\text{XXX}} {x}^{6} - 27 = {k}^{3} - {3}^{3}$
and using the "difference of cubes"
$\textcolor{w h i t e}{\text{XXXXXXX}} = \left(k - 3\right) \left({k}^{2} + 3 k + 9\right)$

Note by checking the discriminant we can see that $\left({k}^{2} + 3 k + 9\right)$ has no real roots (and thus no real factors).

However since $\left(k - 3\right) = \left({x}^{2} - 3\right)$ we can factor, using the "difference of squares" to get
$\textcolor{w h i t e}{\text{XXXXXXX}} = \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left({k}^{2} + 3 k + 9\right)$
or
$\textcolor{w h i t e}{\text{XXXXXXX}} = \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left({x}^{4} + 3 {x}^{2} + 9\right)$