Question

porfa , debo factorizar

Answer 1

es una diferencia de cubos perfectos

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

$(\frac{x}{2})^{3} -(\frac{1}{3})^{3} =(\frac{x}{2} -\frac{1}{3}) [(\frac{x}{2})^{2}-( \frac{x}{2})(\frac{1}{3})+(\frac{1}{3})^{2}]$

$=(\frac{x}{2} -\frac{1}{3}) (\frac{x^{2}}{4}- \frac{x}{6}+\frac{1}{9})$