Problem 63 - Algebra and Trigonometry

Question

Rationalize the denominator. (a) \(\frac{1}{\sqrt[3]{4}}\) (b) \(\frac{1}{\sqrt[4]{3}}\) (c) \(\frac{8}{\sqrt[5]{2}}\)

Step-by-Step Solution

Answer

(a) \( \frac{\sqrt[3]{16}}{4} \); (b) \( \frac{\sqrt[4]{27}}{3} \); (c) \( 4\sqrt[5]{16} \).

Step 1: Understand the Problem

The goal is to eliminate the radical from the denominator by rationalizing it. For a cube root, multiply numerator and denominator by a term that will result in the radicand's exponent becoming a multiple of 3. For a fourth root, multiply by terms that make the exponent a multiple of 4.

Step 2: Rationalize Denominator for (a)

Given: \( \frac{1}{\sqrt[3]{4}} \). Multiply both the numerator and the denominator by \( \sqrt[3]{16} \) (since \( 16 \times 4 = 64 \) which is \(4^3\)) to remove the cube root: \[ \frac{1 \cdot \sqrt[3]{16}}{\sqrt[3]{4} \cdot \sqrt[3]{16}} = \frac{\sqrt[3]{16}}{\sqrt[3]{64}} = \frac{\sqrt[3]{16}}{4} \].

Step 3: Rationalize Denominator for (b)

Given: \( \frac{1}{\sqrt[4]{3}} \). Multiply both the numerator and the denominator by \( \sqrt[4]{27} \) (since \( 27 \times 3 = 81 \) which is \(3^4\)) to remove the fourth root: \[ \frac{1 \cdot \sqrt[4]{27}}{\sqrt[4]{3} \cdot \sqrt[4]{27}} = \frac{\sqrt[4]{27}}{\sqrt[4]{81}} = \frac{\sqrt[4]{27}}{3} \].

Step 4: Rationalize Denominator for (c)

Given: \( \frac{8}{\sqrt[5]{2}} \). Multiply both the numerator and the denominator by \( \sqrt[5]{16} \) (since \( 16 \times 2 = 32 \) which is \(2^5\)) to remove the fifth root: \[ \frac{8 \cdot \sqrt[5]{16}}{\sqrt[5]{2} \cdot \sqrt[5]{16}} = \frac{8\sqrt[5]{16}}{\sqrt[5]{32}} = \frac{8\sqrt[5]{16}}{2} = 4\sqrt[5]{16} \].

Key Concepts

Cube Roots

Rationalizing the denominator with cube roots might seem tricky at first, but it follows a straightforward process. Imagine you have a fraction like \( \frac{1}{\sqrt[3]{4}} \). The goal here is to get rid of the cube root from the denominator, which involves making the radicand in the denominator reach a perfect cube.

• Find the number you need to multiply by, to make the denominator a perfect cube. For example, in \( \sqrt[3]{4} \), you should identify \( 16 \) because when multiplied, \( 4 \times 16 = 64 = 4^3 \).
• Multiply the numerator and the denominator by this number's cube root, \( \sqrt[3]{16} \), which leads to \( \frac{\sqrt[3]{16}}{\sqrt[3]{64}} \).
• Simplify by understanding that \( \sqrt[3]{64} = 4 \), hence the final result is \( \frac{\sqrt[3]{16}}{4} \).

By rationalizing this way, you have successfully removed the cube root from the denominator and simplified your expression!

Fourth Roots

Dealing with fourth roots can feel challenging, but the method is similar to that used for cube roots. Consider the expression \( \frac{1}{\sqrt[4]{3}} \). Here, you need to adjust the denominator so it becomes an exponent of four.

• Choose a multiplication factor that will make the radicand in the denominator a fourth power. For instance, \( 27 \) is suitable here since \( 27 \times 3 = 81 = 3^4 \).
• Multiply both parts of the fraction by this fourth root, \( \sqrt[4]{27} \), resulting in \( \frac{\sqrt[4]{27}}{\sqrt[4]{81}} \).
• This simplifies to \( \frac{\sqrt[4]{27}}{3} \) because \( \sqrt[4]{81} = 3 \).

Through these steps, you effectively eliminate the fourth root from the denominator, cleaning up your mathematical expression.

Fifth Roots

When it comes to rationalizing the denominator with fifth roots, there is an interesting pattern. Taking the example \( \frac{8}{\sqrt[5]{2}} \), our task is to alter the denominator to become a fifth power.

• Locate a number to multiply so that the denominator is raised to a power of five. In this scenario, \( 16 \) is your number, as \( 16 \times 2 = 32 = 2^5 \).
• Multiply the numerator and the denominator by \( \sqrt[5]{16} \), resulting in \( \frac{8\sqrt[5]{16}}{\sqrt[5]{32}} \).
• This simplifies to \( \frac{8\sqrt[5]{16}}{2} \). Eventually, divide 8 by 2 to get \( 4\sqrt[5]{16} \).

By performing these operations, you skillfully remove the fifth root from the denominator, streamlining the overall expression.