\( ( \frac { 2 } { 3 } x ^ { 3 } - \frac { 1 } { y ^ { 2 } } ) ( \frac { 1 } { 2 } x ^ { 2 } - \frac { 4 } { 3 } x y - \frac { 1 } { 2 } y ^ { 2 } ) ( y \neq 0 ) \)

Certainly! Let's expand the given expression step by step. **Given Expression:** \[ \left( \frac{2}{3}x^3 - \frac{1}{y^2} \right) \left( \frac{1}{2}x^2 - \frac{4}{3}xy - \frac{1}{2}y^2 \right) \quad \text{where } y \neq 0 \] **Step 1: Distribute Each Term in the First Parenthesis with Each Term in the Second Parenthesis** We'll use the distributive property (also known as the FOIL method for binomials) to multiply each term in the first polynomial by each term in the second polynomial. \[ \begin{align*} & \left( \frac{2}{3}x^3 \right) \left( \frac{1}{2}x^2 \right) + \left( \frac{2}{3}x^3 \right) \left( -\frac{4}{3}xy \right) + \left( \frac{2}{3}x^3 \right) \left( -\frac{1}{2}y^2 \right) \\ & + \left( -\frac{1}{y^2} \right) \left( \frac{1}{2}x^2 \right) + \left( -\frac{1}{y^2} \right) \left( -\frac{4}{3}xy \right) + \left( -\frac{1}{y^2} \right) \left( -\frac{1}{2}y^2 \right) \end{align*} \] **Step 2: Multiply the Terms** \[ \begin{align*} & \frac{2}{3} \times \frac{1}{2} x^{3+2} + \frac{2}{3} \times -\frac{4}{3} x^{3+1}y + \frac{2}{3} \times -\frac{1}{2} x^3 y^2 \\ & + -\frac{1}{y^2} \times \frac{1}{2}x^2 + -\frac{1}{y^2} \times -\frac{4}{3}xy + -\frac{1}{y^2} \times -\frac{1}{2}y^2 \\ = \quad & \frac{1}{3}x^5 - \frac{8}{9}x^4y - \frac{1}{3}x^3y^{-2} - \frac{1}{2}x^2y^{-2} + \frac{4}{3}xy^{-1} + \frac{1}{2}y^{-4} \end{align*} \] **Step 3: Simplify the Expression** Combining all the terms, the expanded form of the original expression is: \[ \frac{1}{3}x^5 - \frac{8}{9}x^4y - \frac{1}{3}\frac{x^3}{y^2} - \frac{1}{2}\frac{x^2}{y^2} + \frac{4}{3}\frac{x}{y} + \frac{1}{2}\frac{1}{y^4} \] **Final Expanded Expression:** \[ \boxed{\,\frac{1}{3}x^{5} - \frac{8}{9}x^{4}y - \frac{1}{3}\frac{x^{3}}{y^{2}} - \frac{1}{2}\frac{x^{2}}{y^{2}} + \frac{4}{3}\frac{x}{y} + \frac{1}{2}\frac{1}{y^{4}}\,} \]