How do you rationalize the denominator and simplify $\sqrt[3]{\frac{9}{7}}$ $root3(9/7)$ ?

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How do you rationalize the denominator and simplify $\sqrt[3]{\frac{9}{7}}$ $root3(9/7)$ ?

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Answer 1 · 2016-03-15T18:44:37

$\frac{\sqrt[3]{441}}{7}$ $root(3)(441)/7$

Explanation:

$1$ $1$ . Since the denominator contains a cube root, we need to multiply the numerator and denominator by a value that will result in the denominator of $\sqrt[3]{\frac{9}{7}}$ $root(3)(9/7)$ to have a perfect cube. Thus, start by multiplying the numerator and denominator by $\sqrt[3]{49}$ $root(3)(49)$ .

$\sqrt[3]{\frac{9}{7}}$ $root(3)(9/7)$

$= \frac{\sqrt[3]{9}}{\sqrt[3]{7}}$ $=root(3)(9)/root(3)(7)$

$= \frac{\sqrt[3]{9}}{\sqrt[3]{7}} (\frac{\sqrt[3]{49}}{\sqrt[3]{49}})$ $=root(3)(9)/root(3)(7)(root(3)(49)/root(3)(49))$

$2$ $2$ . Simplify.

$= \frac{\sqrt[3]{441}}{\sqrt[3]{343}}$ $=root(3)(441)/root(3)(343)$

$=color(green)(|bar(ul(color(white)(a/a)root(3)(441)/7color(white)(a/a)|)))$

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How do you rationalize the denominator and simplify $\sqrt[3]{\frac{9}{7}}$ $root3(9/7)$ ?

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