which of the following is equivalent to \frac{\sqrt3{32x^3y^6}}{\sqrt3{2x^9y^2}}, where $x \ge 0$ and $y \ge 0$?

* $\sqrt3{16x^6y^4}$
* $\sqrt3{\frac{y^4}{16x^6}}$
* $\sqrt3{\frac{16y^4}{x^6}}$

Question

which of the following is equivalent to \frac{\sqrt3{32x^3y^6}}{\sqrt3{2x^9y^2}}, where $x \ge 0$ and $y \ge 0$?

* $\sqrt3{16x^6y^4}$
* $\sqrt3{\frac{y^4}{16x^6}}$
* $\sqrt3{\frac{16y^4}{x^6}}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Exponent Rules", "Algebraic Simplification" ], "new_concepts": [], "current_concepts": [ "Exponent Rules", "Algebraic Simplification", "Radical Simplification" ] } ## Apply quotient property of radicals Using the *Algebraic Simplification* knowledge point $$ \frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2}} = \sqrt[3]{\frac{32x^3y^6}{2x^9y^2}} $$ ## Simplify the fraction inside the radical Using the *Exponent Rules* knowledge point $$ \begin{aligned} \frac{32}{2} &= 16 \ \frac{x^3}{x^9} &= x^{3-9} = x^{-6} = \frac{1}{x^6} \ \frac{y^6}{y^2} &= y^{6-2} = y^4 \ \frac{32x^3y^6}{2x^9y^2} &= \frac{16y^4}{x^6} \end{aligned} $$ ## Write the final simplified radical expression Using the *Algebraic Simplification* knowledge point $$ \sqrt[3]{\frac{16y^4}{x^6}} $$

(A) $\sqrt[3]{16x^6y^4}$

(B) $\sqrt[3]{\frac{y^4}{16x^6}}$

(C) $\sqrt[3]{\frac{16y^4}{x^6}}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Radical Simplification" ] }

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QUESTION IMAGE

Question

Explanation:

Apply quotient property of radicals

Simplify the fraction inside the radical

Write the final simplified radical expression

Answer: