Answer
$=-x\sqrt [3] {2xy}$.
Work Step by Step
The given expression is
$=\frac{16x^4\sqrt [3] {48x^3y^2}}{8x^3\sqrt [3]{3x^2y}}-\frac{20\sqrt [3]{2x^3y}}{4\sqrt[3] {x^{-1}}}$
Divide the radicands and retain the common index.
$=\frac{16x^4}{8x^3}\cdot\sqrt [3] {\frac{48x^3y^2}{3x^2y}}-\frac{20}{4}\sqrt [3]{\frac{2x^3y}{x^{-1}}}$
Divide factors. Subtract exponents on common bases.
$=2x^{4-3}\cdot\sqrt [3] {16x^{3-2}y^{2-1}}-5\sqrt [3]{2x^{3+1}y}$
Simplify.
$=2x^{1}\cdot\sqrt [3] {16x^{1}y^{1}}-5\sqrt [3]{2x^{4}y}$
$=2x\cdot2\sqrt [3] {2xy}-5x\sqrt [3]{2xy}$
Simplify.
$=4x\sqrt [3] {2xy}-5x\sqrt [3]{2xy}$
Apply the distributive property.
$=(4x-5x)\sqrt [3] {2xy}$
Simplify.
$=-x\sqrt [3] {2xy}$.
