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

May 21, 2015

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

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

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

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

This is of the form $\left({a}^{3} - {b}^{3}\right)$ which has a well known factorization:

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

So we can substitute $a = \frac{x}{2}$, $b = \frac{y}{3}$ to get

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

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

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

This is as far as we can go with real valued coefficients.