# How do you factor 216m^3+1/64?

Jun 2, 2015

$216 {m}^{3} + \frac{1}{64}$

$= {\left(6 m\right)}^{3} + {\left(\frac{1}{4}\right)}^{3}$

$= \left(\left(6 m\right) + \left(\frac{1}{4}\right)\right) \left({\left(6 m\right)}^{2} - \left(6 m\right) \left(\frac{1}{4}\right) + {\left(\frac{1}{4}\right)}^{2}\right)$

$= \left(6 m + \frac{1}{4}\right) \left(36 {m}^{2} - \frac{3 m}{2} + \frac{1}{16}\right)$

...using the sum of cubes identity

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

There are no simpler factors with real coefficients.