Q. 263 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator.  

$$\begin{gathered} a) \frac{1}{\sqrt[3]{5}} \\ b) \sqrt[3]{\frac{5}{24}} \\ c) \frac{4}{\sqrt[3]{36a}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[3]{25}}{5}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[3]{45}}{6}$.

Part (c) The rationalization of the denominator is $\frac{2\sqrt[3]{6a^2}}{3a}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[3]{5}} \\ b) \sqrt[3]{\frac{5}{24}} \\ c) \frac{4}{\sqrt[3]{36a}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{5^2}$.

$\frac{1}{\sqrt[3]{5}}=\frac{1\times \sqrt[3]{5^2}}{\sqrt[3]{5}\times \sqrt[3]{5^2}}$

On simplifying we get,

$\frac{1\times \sqrt[3]{5^2}}{\sqrt[3]{5}\times \sqrt[3]{5^2}}=\frac{\sqrt[3]{25}}{5}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{25}}{5}$.

Step 3 Solving part (b).

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

$$\begin{gathered} \sqrt[3]{\frac{5}{24}}=\frac{\sqrt[3]{5}}{\sqrt[3]{24}} \\ =\frac{\sqrt[3]{5}}{2\sqrt[3]{3}} \end{gathered}$$

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

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

Then on simplifying we get,

$\frac{\sqrt[3]{5}\times \sqrt[3]{3^2}}{2\sqrt[3]{3}\times \sqrt[3]{3^2}}=\frac{\sqrt[3]{45}}{6}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{45}}{6}$.

Step 4. Solving Part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{{\left(36\right)}^2{a}^2}$.

$\frac{4}{\sqrt[3]{36a}}=\frac{4\times \sqrt[3]{{\left(36\right)}^2{a}^2}}{\sqrt[3]{36a}\times \sqrt[3]{{\left(36\right)}^2{a}^2}}$

On simplifying we get,

$\frac{4\times \sqrt[3]{{\left(36\right)}^2{a}^2}}{\sqrt[3]{36a}\times \sqrt[3]{{\left(36\right)}^2{a}^2}}=\frac{24\sqrt[3]{6a^2}}{36a}$

Then removing the common factors.

$\frac{24\sqrt[3]{6a^2}}{36a}=\frac{2\sqrt[3]{6a^2}}{3a}$

Hence, the rationalization of the denominator is $\frac{2\sqrt[3]{6a^2}}{3a}$.