Q. 266 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[3]{13}} \\ b) \sqrt[3]{\frac{3}{128}} \\ c) \frac{3}{\sqrt[3]{6y^2}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[3]{169}}{13}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[3]{12}}{8}$.

Part (c) The rationalization of the denominator is $\frac{\sqrt[3]{36y}}{2y}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[3]{13}} \\ b) \sqrt[3]{\frac{3}{128}} \\ c) \frac{3}{\sqrt[3]{6y^2}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{{\left(13\right)}^2}$.

$\frac{1}{\sqrt[3]{13}}=\frac{1\times \sqrt[3]{{\left(13\right)}^2}}{\sqrt[3]{13}\times \sqrt[3]{{\left(13\right)}^2}}$

On simplifying we get,

$\frac{1\times \sqrt[3]{{\left(13\right)}^2}}{\sqrt[3]{13}\times \sqrt[3]{{\left(13\right)}^2}}=\frac{\sqrt[3]{169}}{13}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{169}}{13}$.

Step 3. Solving part (b).

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

$$\begin{gathered} \sqrt[3]{\frac{3}{128}}=\frac{\sqrt[3]{3}}{\sqrt[3]{128}} \\ =\frac{\sqrt[3]{3}}{4\sqrt[3]{2}} \end{gathered}$$

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

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

Then on simplifying we get,

$\frac{\sqrt[3]{3}\times \sqrt[3]{2^2}}{4\sqrt[3]{2}\times \sqrt[3]{2^2}}=\frac{\sqrt[3]{12}}{8}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{12}}{8}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{6^2{y}^4}$.

$\frac{3}{\sqrt[3]{6y^2}}=\frac{3\times \sqrt[3]{6^2{y}^4}}{\sqrt[3]{6y^2}\times \sqrt[3]{6^2{y}^4}}$

On simplifying we get,

$\frac{3\times \sqrt[3]{6^2{y}^4}}{\sqrt[3]{6y^2}\times \sqrt[3]{6^2{y}^4}}=\frac{3y\sqrt[3]{36y}}{6y^2}$

Then on removing the common factors.

$\frac{3y\sqrt[3]{36y}}{6y^2}=\frac{\sqrt[3]{36y}}{2y}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{36y}}{2y}$.