How do you rationalize the denominator and simplify $\frac{1}{\sqrt[3]{x^{2}}}$ $1/root3(x^2)$ ?

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How do you rationalize the denominator and simplify $\frac{1}{\sqrt[3]{x^{2}}}$ $1/root3(x^2)$ ?

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Answer 1 · 2017-10-17T12:01:22

$\frac{1}{\sqrt[3]{x^{2}}} = x^{- \frac{4}{3}}$ $1/root(3)(x^2)=color(red)(x^(-4/3))$

Explanation:

$\frac{1}{\sqrt[3]{x^{2}}}$ $1/root(3)(x^2)$
$XXX = \frac{1}{\sqrt[3]{x^{2}}} \times \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}} \times \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}$ $color(white)("XXX")=1/root(3)(x^2)xxroot(3)(x^2)/root(3)(x^2)xxroot(3)(x^2)/root(3)(x^2)$

$XXX = \frac{{(\sqrt[3]{x^{2}})}^{2}}{x^{2}}$ $color(white)("XXX")=((root(3)(x^2))^2)/(x^2)$

$XXX = {(x^{\frac{1}{3}})}^{2} \times ` \frac{1}{x^{2}}$ $color(white)("XXX")=(x^(1/3))^2xx`1/(x^2)$

$XXX = x^{\frac{2}{3}} \times x^{- 2}$ $color(white)("XXX")=x^(2/3) xx x^(-2)$

$XXX = x^{(\frac{2}{3} - 2)}$ $color(white)("XXX")=x^((2/3-2))$

$XXX = x^{(- \frac{4}{3})}$ $color(white)("XXX")=x^((-4/3))$

Answer 2 · 2017-10-17T12:03:38

See a solution process below:

Explanation:

To rationalize the denominator we must multiply the fraction by the appropriate value of $1$ $1$ :

$\frac{\sqrt[3]{x}}{\sqrt[3]{x}} \times \frac{1}{\sqrt[3]{x^{2}}} \Rightarrow$ $root(3)(x)/root(3)(x) xx 1/root(3)(x^2) =>$

$\frac{\sqrt[3]{x} \times 1}{\sqrt[3]{x} \times \sqrt[3]{x^{2}}} \Rightarrow$ $(root(3)(x) xx 1)/(root(3)(x) xx root(3)(x^2)) =>$

$\frac{\sqrt[3]{x}}{\sqrt[3]{x \times x^{2}}} \Rightarrow$ $root(3)(x)/(root(3)(x xx x^2)) =>$

$\frac{\sqrt[3]{x}}{\sqrt[3]{x^{3}}} \Rightarrow$ $root(3)(x)/(root(3)(x^3)) =>$

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

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