# How do you factor 27x^3-(1/8)?

May 9, 2015

This is an example of the difference of cubes, in which $\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ .

Factor $27 {x}^{3} - \frac{1}{8}$ .

$a = 3 x$
$b = \frac{1}{2}$

$\left(3 x - \frac{1}{2}\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(\frac{1}{2}\right) + {\left(\frac{1}{2}\right)}^{2}\right)$ =

$\left(3 x - \frac{1}{2}\right) \left(9 {x}^{2} + \frac{3 x}{2} + \frac{1}{4}\right)$

May 9, 2015

Notice that $27$, ${x}^{3}$ and $\frac{1}{8}$ are all cubes. If you were allowed complex numbers as coefficients, then this could be completely factored as $\left(3 x - \frac{1}{2}\right) \left(3 \omega x - \frac{1}{2}\right) \left(3 {\omega}^{2} x - \frac{1}{2}\right)$, where $\omega$ is the complex cube root of 1.

As it is, you are only interested in real numbers, not complex ones. So the only linear factor is $\left(3 x - \frac{1}{2}\right)$. The other factor is $\left(9 {x}^{2} + \left(\frac{3}{2}\right) x + \frac{1}{4}\right)$.

Here are a couple of ways of finding that quadratic factor:

(1) Using complex arithmetic:
$\left(3 \omega x - \frac{1}{2}\right) \left(3 {\omega}^{2} x - \frac{1}{2}\right) = 9 {\omega}^{3} {x}^{2} - \left(\frac{3}{2}\right) \left(\omega + {\omega}^{2}\right) x + \frac{1}{4} = 9 {x}^{2} + \left(\frac{3}{2}\right) x + \frac{1}{4}$
(since ${\omega}^{3} = 1$ and $\omega + {\omega}^{2} = - 1$).

(2) Using real arithmetic, solve:
$27 {x}^{3} - \frac{1}{8} = \left(3 x - \frac{1}{2}\right) \left(a {x}^{2} + b x + c\right)$
$= 3 a {x}^{3} + \left(3 b - \left(\frac{1}{2}\right) a\right) {x}^{2} + \left(3 c - \left(\frac{1}{2}\right) b\right) x - \left(\frac{1}{2}\right) c$
Comparing coefficients, we get:
$3 a = 27$
$\left(3 b - \left(\frac{1}{2}\right) a\right) = 0$
$\left(3 c - \left(\frac{1}{2}\right) b\right) = 0$
$\left(\frac{1}{2}\right) c = \frac{1}{8}$
yielding $a = 9$, $b = \frac{3}{2}$ and $c = \frac{1}{4}$.