Question #79897

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Question #79897

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Answer 1 · 2016-10-16T22:22:53

$\frac{\sqrt[3]{7 x}}{\sqrt[3]{3 y}} = \frac{\sqrt[3]{63 x y^{2}}}{3 y}$ $root(3)(7x)/root(3)(3y)=root(3)(63xy^2)/(3y)$

Explanation:

Assuming "rationalize" refers to rationalizing the denominator , note that $\sqrt[3]{3 y}$ $root(3)(3y)$ is the value such that ${(\sqrt[3]{3 y})}^{3} = 3 y$ $(root(3)(3y))^3 = 3y$ . We will use that fact to eliminate the root from the denominator.

$\frac{\sqrt[3]{7 x}}{\sqrt[3]{3 y}} = \frac{\sqrt[3]{7 x} \cdot {(\sqrt[3]{3 y})}^{2}}{\sqrt[3]{3 y} \cdot {(\sqrt[3]{3 y})}^{2}}$ $root(3)(7x)/root(3)(3y) = (root(3)(7x) * (root(3)(3y))^2) / (root(3)(3y) * (root(3)(3y))^2)$

$= \frac{\sqrt[3]{7 x} \cdot \sqrt[3]{{(3 y)}^{2}}}{{(\sqrt[3]{3 y})}^{3}}$ $=(root(3)(7x) * root(3)((3y)^2))/(root(3)(3y))^3$

$= \frac{\sqrt[3]{7 x \cdot {(3 y)}^{2}}}{3 y}$ $=root(3)(7x*(3y)^2)/(3y)$

$= \frac{\sqrt[3]{63 x y^{2}}}{3 y}$ $=root(3)(63xy^2)/(3y)$

(Note that we also used the properties ${(\sqrt[a]{x})}^{b} = \sqrt[a]{x^{b}}$ $(root(a)(x))^b = root(a)(x^b)$ and $\sqrt[a]{x} \cdot \sqrt[a]{y} = \sqrt[a]{x y}$ $root(a)(x)*root(a)(y) = root(a)(xy)$ during the process)

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Question #79897

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