Question
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Given \( x>0 \) and \( y>0 \), select the expression that is equivalent to \[ \sqrt[3]{-64 x^{9} y^{2}} \] Answer \( \begin{array}{ll}\text { ( } 4 i x^{3} y^{\frac{2}{3}} & -4 x^{\frac{1}{3}} y^{\frac{3}{2}}\end{array} \) Nth

Ask by Tyler Ford. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The expression equivalent to \(\sqrt[3]{-64x^{9}y^{2}}\) is \(-4x^{3}\sqrt[3]{y^{2}}\).

Solution

Calculate or simplify the expression \( \sqrt[3]{-64x^9y^2} \). Simplify the expression by following steps: - step0: Solution: \(\sqrt[3]{-64x^{9}y^{2}}\) - step1: Use the properties of radicals: \(-\sqrt[3]{64x^{9}y^{2}}\) - step2: Simplify the expression: \(-4x^{3}\sqrt[3]{y^{2}}\) The expression equivalent to \(\sqrt[3]{-64x^{9}y^{2}}\) is \(-4x^{3}\sqrt[3]{y^{2}}\).

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Beyond the Answer

First, let’s simplify the expression \( \sqrt[3]{-64 x^{9} y^{2}} \). The cube root of a negative number is expressed as \( - \) times the cube root of the positive version. Thus, we have: \[ \sqrt[3]{-64 x^{9} y^{2}} = -\sqrt[3]{64} \cdot \sqrt[3]{x^{9}} \cdot \sqrt[3]{y^{2}} = -4 x^{3} y^{\frac{2}{3}}. \] Now, looking closely, \( 4i x^{3} y^{\frac{2}{3}} \) is incorrect because it introduces an imaginary unit \( i \) which does not apply here; the cube root of a negative real number results in a negative real value, so the simplified expression is indeed: \(-4 x^{3} y^{\frac{2}{3}}.\) This ensures that the expression maintains its negative aspect without introducing imaginary numbers. So, the equivalent expression is: \(-4 x^{3} y^{\frac{2}{3}}.\)

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