How do you rationalize the denominator and simplify $\frac{12}{\sqrt[3]{9}}$ $12/root3(9)$ ?

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Answer 1 · 2018-04-16T21:19:06

$4 \sqrt[3]{3}$ $4root3(3)$

$\frac{12}{\sqrt[3]{9}}$ $12/(root3(9))$

To rationalize, multiply the expression by $\frac{\sqrt[3]{9^{2}}}{\sqrt[3]{9^{2}}}$ $root3(9^2)/root3(9^2)$ :

$\frac{12}{\sqrt[3]{9}} \cdot \frac{\sqrt[3]{9^{2}}}{\sqrt[3]{9^{2}}} =$ $12/(root3(9))*root3(9^2)/root3(9^2)=$

$\frac{12 (\sqrt[3]{9^{2}})}{\sqrt[3]{9^{3}}} =$ $(12(root3(9^2)))/root3(9^3)=$

$\frac{12 (\sqrt[3]{81})}{9} =$ $(12(root3(81)))/9=$

$\frac{12 (\sqrt[3]{27 \cdot 3})}{9} =$ $(12(root3(27*3)))/9=$

$\frac{12 \cdot 3 (\sqrt[3]{3})}{9} =$ $(12*3(root3(3)))/9=$

$\frac{36 (\sqrt[3]{3})}{9} =$ $(36(root3(3)))/9=$

$4 \sqrt[3]{3}$ $4root3(3)$

How do you rationalize the denominator and simplify 123√912/root3(9)?