Q. 264 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator.  

$$\begin{gathered} a) \frac{1}{\sqrt[3]{3}} \\ b) \sqrt[3]{\frac{5}{32}} \\ c) \frac{7}{\sqrt[3]{49b}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[3]{9}}{3}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[3]{10}}{4}$.

Part (c) The rationalization of the denominator is $\frac{\sqrt[3]{7b^2}}{b}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[3]{3}} \\ b) \sqrt[3]{\frac{5}{32}} \\ c) \frac{7}{\sqrt[3]{49b}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{3^2}$.

$\frac{1}{\sqrt[3]{3}}=\frac{1\times \sqrt[3]{3^2}}{\sqrt[3]{3}\times \sqrt[3]{3^2}}$

On simplifying we get

$\frac{1\times \sqrt[3]{3^2}}{\sqrt[3]{3}\times \sqrt[3]{3^2}}=\frac{\sqrt[3]{9}}{3}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{9}}{3}$.

Step 2. Solving part (b).

Now, to rationalize the expression first we will rewrite the expression using quotient property and simplify. 

$$\begin{gathered} \sqrt[3]{\frac{5}{32}}=\frac{\sqrt[3]{5}}{\sqrt[3]{32}} \\ =\frac{\sqrt[3]{5}}{2\sqrt[3]{4}} \end{gathered}$$

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

$\frac{\sqrt[3]{5}}{2\sqrt[3]{4}}=\frac{\sqrt[3]{5}\times \sqrt[3]{4^2}}{2\sqrt[3]{4}\times \sqrt[3]{4^2}}$

Then on simplifying we get,

$\frac{\sqrt[3]{5}\times \sqrt[3]{4^2}}{2\sqrt[3]{4}\times \sqrt[3]{4^2}}=\frac{2\sqrt[3]{10}}{8}$

Finally removing the common factors.

$\frac{2\sqrt[3]{10}}{8}=\frac{\sqrt[3]{10}}{4}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{10}}{4}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{{\left(49\right)}^2{b}^2}$.

$\frac{7}{\sqrt[3]{49b}}=\frac{7\times \sqrt[3]{{\left(49\right)}^2{b}^2}}{\sqrt[3]{49b}\times \sqrt[3]{{\left(49\right)}^2{b}^2}}$

On simplifying we get,

$\frac{7\times \sqrt[3]{{\left(49\right)}^2{b}^2}}{\sqrt[3]{49b}\times \sqrt[3]{{\left(49\right)}^2{b}^2}}=\frac{49\sqrt[3]{7b^2}}{49b}$

Then removing the common factors.

$\frac{49\sqrt[3]{7b^2}}{49b}=\frac{\sqrt[3]{7b^2}}{b}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{7b^2}}{b}$.