Context: For positive $x$ and $y$ we must decide which of forms I and II are equivalent to $\dfrac{\sqrt[3]{x^{17}y^5}}{5x^2(\sqrt[8]{y^8})}$.

Answer: C) I and II

Why: Since $\sqrt[8]{y^8} = y$ for positive $y$, the expression equals $\dfrac{x^{17/3}y^{5/3}}{5x^2y} = \dfrac{x^{17/3 - 2}y^{5/3 - 1}}{5} = \dfrac{x^{11/3}y^{2/3}}{5}$. Form I: $\dfrac{x^{14/3}y^{5/3}}{5xy} = \dfrac{x^{14/3 - 1}y^{5/3 - 1}}{5} = \dfrac{x^{11/3}y^{2/3}}{5}$ — equivalent. Form II: $\dfrac{\sqrt[3]{x^{11}y^2}}{5} = \dfrac{x^{11/3}y^{2/3}}{5}$ — also equivalent.

Traps:

• A) I only — wrongly rejects II, but $\sqrt[3]{x^{11}y^2}$ is exactly $x^{11/3}y^{2/3}$, the fully simplified form.
• B) II only — wrongly rejects I, but dividing $x^{14/3}$ by $x$ and $y^{5/3}$ by $y$ lands on the same powers $x^{11/3}y^{2/3}$.
• D) Neither — both forms reduce to $\dfrac{x^{11/3}y^{2/3}}{5}$; choosing this reflects an exponent-subtraction slip (e.g. mishandling $\dfrac{17}{3} - 2 = \dfrac{11}{3}$).