Q. 265 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[3]{11}} \\ b) \sqrt[3]{\frac{7}{54}} \\ c) \frac{3}{\sqrt[3]{3x^2}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[3]{121}}{11}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[3]{28}}{6}$.

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

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[3]{11}} \\ b) \sqrt[3]{\frac{7}{54}} \\ c) \frac{3}{\sqrt[3]{3x^2}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[3]{{\left(11\right)}^2}$.

$\frac{1}{\sqrt[3]{11}}=\frac{1\times \sqrt[3]{{\left(11\right)}^2}}{\sqrt[3]{11}\times \sqrt[3]{{\left(11\right)}^2}}$

On simplifying we get,

$\frac{1\times \sqrt[3]{{\left(11\right)}^2}}{\sqrt[3]{11}\times \sqrt[3]{{\left(11\right)}^2}}=\frac{\sqrt[3]{121}}{11}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{121}}{11}$.

Step 3. Solving part (b).

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

$$\begin{gathered} \sqrt[3]{\frac{7}{54}}=\frac{\sqrt[3]{7}}{\sqrt[3]{54}} \\ =\frac{\sqrt[3]{7}}{3\sqrt[3]{2}} \end{gathered}$$

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

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

Then on simplifying we get,

$\frac{\sqrt[3]{7}\times \sqrt[3]{2^2}}{3\sqrt[3]{2}\times \sqrt[3]{2^2}}=\frac{\sqrt[3]{28}}{6}$

Hence, the rationalization of the denominator is $\frac{\sqrt[3]{28}}{6}$.

Step 4. Solving part (c).

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

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

On simplifying we get,

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

Then removing the common factors.

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

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