Answer
$\dfrac{\sqrt[3]{128x^{3}}}{-3\sqrt[3]{2x}}=-\dfrac{4}{3}\sqrt[3]{x^{2}}$
Work Step by Step
$\dfrac{\sqrt[3]{128x^{3}}}{-3\sqrt[3]{2x}}$
Rewrite this expression as $-\dfrac{1}{3}\sqrt[3]{\dfrac{128x^{3}}{2x}}$
$\dfrac{\sqrt[3]{128x^{3}}}{-3\sqrt[3]{2x}}=-\dfrac{1}{3}\sqrt[3]{\dfrac{128x^{3}}{2x}}=...$
Evaluate the division inside the cubic root and simplify:
$...=-\dfrac{1}{3}\sqrt[3]{64x^{2}}=-\dfrac{1}{3}\cdot\sqrt[3]{64}\cdot\sqrt[3]{x^{2}}=-\dfrac{1}{3}\cdot4\cdot\sqrt[3]{x^{2}}=-\dfrac{4}{3}\sqrt[3]{x^{2}}$
