Answer
$\sqrt [3]{3x}/6$
Work Step by Step
$\sqrt [3]{24x/27} - \sqrt [3]{3x}/2$
$\sqrt [3]{24x/3^3}- \sqrt [3]{3x}/2$
$\sqrt [3]{2^3*3x}/3- \sqrt [3]{3x}/2$
$2\sqrt [3]{3x}/3- \sqrt [3]{3x}/2$
$2*2\sqrt [3]{3x}/3*2- 3*\sqrt [3]{3x}/2*3$
$4*\sqrt [3]{3x}/6- 3*\sqrt [3]{3x}/6$
$\sqrt [3]{3x}/6$
