In the following exercises, rationalize the denominator.

$$\begin{gathered} a)\frac{1}{\sqrt[4]{4}} \\ b)\sqrt[4]{\frac{9}{32}} \\ c)\frac{6}{\sqrt[4]{9x^3}} \end{gathered}$$

We have,

$$\begin{gathered} a)\frac{1}{\sqrt[4]{4}} \\ b)\sqrt[4]{\frac{9}{32}} \\ c)\frac{6}{\sqrt[4]{9x^3}} \end{gathered}$$

To rationalize the denominator we will multiply numerator and denominator by the conjugate of the denominator.

Also we will use the radical rule,

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$\sqrt[n]{a^n}=a$

${\left(a^b\right)}^c=a^{bc}$

We have,

$\frac{1}{\sqrt[4]{4}}$

Rewriting it,

role="math" localid="1645710687226" $\frac{1}{\sqrt[4]{4}}=\frac{1}{\sqrt[4]{2^2}}$

$\Rightarrow \frac{1}{\sqrt[4]{4}}=\frac{1}{\sqrt{2}}$

Multiplying numerator and denominator by $\sqrt{2}$,

role="math" localid="1645711062867" $\Rightarrow \frac{1}{\sqrt[4]{4}}=\frac{1·\sqrt{2}}{\sqrt{2}·\sqrt{2}}$

Simplifying it we get,

$\Rightarrow \frac{1}{\sqrt[4]{4}}=\frac{\sqrt{2}}{2}$

Hence, rationalizing $\frac{1}{\sqrt[4]{4}}$we will get$\frac{\sqrt{2}}{2}$.

We have,

$\sqrt[4]{\frac{9}{32}}$

Rewriting it,

$\sqrt[4]{\frac{9}{32}}=\frac{\sqrt[4]{9}}{\sqrt[4]{32}}$

$\Rightarrow \sqrt[4]{\frac{9}{32}}=\frac{\sqrt[4]{3^2}}{\sqrt[4]{2^4·2}}$

Using radical rule,

$\Rightarrow \sqrt[4]{\frac{9}{32}}=\frac{\sqrt{3}}{2\sqrt[4]{2}}$

Multiplying numerator and denominator by $\sqrt[4]{2^3}$,

$\sqrt[4]{\frac{9}{32}}=\frac{\sqrt{3}·\sqrt[4]{2^3}}{2\sqrt[4]{2}·\sqrt[4]{2^3}}$

Using radical rule,

role="math" localid="1645711576856" $\sqrt[4]{\frac{9}{32}}=\frac{\sqrt{3}·\sqrt[4]{2^3}}{4}$

Hence, rationalizing $\sqrt[4]{\frac{9}{32}}$we will get$\frac{\sqrt{3}·\sqrt[4]{2^3}}{4}$.

We have,

$\frac{6}{\sqrt[4]{9x^3}}$

Rewriting it,

$\frac{6}{\sqrt[4]{9x^3}}=\frac{6}{\sqrt[4]{3^2{x}^3}}$

Multiplying numerator and denominator by $\sqrt[4]{3^{2}x}$,

$\frac{6}{\sqrt[4]{9x^3}}=\frac{6·\sqrt[4]{3^{2}x}}{3x}$

$\Rightarrow \frac{6}{\sqrt[4]{9x^3}}=\frac{2\sqrt{3}\sqrt[4]{x}}{x}$

Hence, rationalizing$\frac{6}{\sqrt[4]{9x^3}}$we will get$\frac{2\sqrt{3}\sqrt[4]{x}}{x}$.