How do you factor 108m^3 - 500?

Mar 18, 2018

$= \left(3 m - 5\right) \left(6 m + 5 \left(1 + \sqrt{3} i\right)\right) \left(6 m + 5 \left(1 - \sqrt{3} i\right)\right)$

Explanation:

$108 = 4 \cdot 27 = 4 \cdot {3}^{3}$
$500 = 4 \cdot 125 = 4 \cdot {5}^{3}$
$\text{So we have}$
$4 \left({\left(3 m\right)}^{3} - {5}^{3}\right)$
$\text{Now we apply "a^3-b^3 = (a-b)(a^2+ab+b^2)}$
$= 4 \left(3 m - 5\right) \left(9 {m}^{2} + 15 m + 25\right)$
$\text{The quadratic factor can also be factored in factors}$
$\text{with complex numbers as follows :}$
$\text{disc : } {15}^{2} - 4 \cdot 9 \cdot 25 = - 675 = - 27 \cdot 25 = - 27 \cdot {5}^{2}$
$\implies m = \frac{- 15 \pm 5 \sqrt{27} i}{18}$
$\implies m = \frac{- 5 \pm 5 \sqrt{3} i}{6}$
$\implies m = - \left(\frac{5}{6}\right) \left(1 \pm \sqrt{3} i\right)$
$\implies 9 \left(m + \left(\frac{5}{6}\right) \left(1 + \sqrt{3} i\right)\right) \left(m + \left(\frac{5}{6}\right) \left(1 - \sqrt{3} i\right)\right)$
$\text{So we get}$
$36 \left(3 m - 5\right) \left(m + \left(\frac{5}{6}\right) \left(1 + \sqrt{3} i\right)\right) \left(m + \left(\frac{5}{6}\right) \left(1 - \sqrt{3} i\right)\right)$
$= \left(3 m - 5\right) \left(6 m + 5 \left(1 + \sqrt{3} i\right)\right) \left(6 m + 5 \left(1 - \sqrt{3} i\right)\right)$