Problem 85 - Algebra and Trigonometry

Question

\(83-88=\) Rationalize the denominator. $$ \begin{array}{llll}{\text { (a) } \frac{1}{\sqrt[3]{4}}} & {\text { (b) } \frac{1}{\sqrt[4]{3}}} & {\text { (c) } \frac{8}{\sqrt[5]{2}}}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Rationalizing the denominator for (a)

We have \(\frac{1}{\sqrt[3]{4}}\). To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt[3]{16}\), because \(4 \times 16 = 64\), which is a perfect cube. This gives \(\frac{\sqrt[3]{16}}{\sqrt[3]{64}} = \frac{\sqrt[3]{16}}{4}\).

Step 2: Rationalizing the denominator for (b)

We have \(\frac{1}{\sqrt[4]{3}}\). Multiply the numerator and the denominator by \(\sqrt[4]{27}\), because \(3 \times 27 = 81\), which is a perfect fourth power. This results in \(\frac{\sqrt[4]{27}}{\sqrt[4]{81}} = \frac{\sqrt[4]{27}}{3}\).

Step 3: Rationalizing the denominator for (c)

We have \(\frac{8}{\sqrt[5]{2}}\). To rationalize, multiply the numerator and the denominator by \(\sqrt[5]{16}\), since \(2 \times 16 = 32\), a perfect fifth power. This leads us to \(\frac{8\cdot \sqrt[5]{16}}{2} = 4\cdot \sqrt[5]{16}\).

Key Concepts

Perfect Cube

Understanding the concept of a perfect cube is important when dealing with cube roots in mathematics. A perfect cube is an integer that is the cube of another integer. In simpler terms, it's a number that can be expressed as the product of three identical factors of a number. For example, 27 is a perfect cube because it can be expressed as

• \[ 3 imes 3 imes 3 = 27 \]
• which is \(3^3\).

When simplifying expressions like \(\frac{1}{\sqrt[3]{4}}\), we aim to remove the cube root in the denominator. To do this, find a number that multiplied by 4 becomes a perfect cube. In this case, it's 16, because

• \[ 4 \times 16 = 64 \]
• and 64 is \(4^3\).

Thus, multiply both the numerator and the denominator by \(\sqrt[3]{16}\) to rationalize the cube root.

Perfect Fourth Power

Next, let’s dive into the perfect fourth power concept, especially as seen in fourth roots. A perfect fourth power is an integer that results from raising a number to the power of four. It is like squaring a square. For example, 16 is a perfect fourth power because

• \[ 2^4 = 16 \]
• or \(2 \times 2 \times 2 \times 2\).

In exercises like \(\frac{1}{\sqrt[4]{3}}\), we aim to find a number that makes 3 into a perfect fourth power when multiplied. This number is 27, as

• \[ 3 \times 27 = 81 \]
• where 81 is \(3^4\).

So, multiplying both numerator and denominator by \(\sqrt[4]{27}\) helps eliminate the fourth root from the denominator.

Perfect Fifth Power

Perfect fifth power is a similar concept, but it relates to raising a number to the power of five. It’s more extensive than the previous cases but follows a similar logic. A number like 32 is a perfect fifth power because

• \[ 2^5 = 32 \]
• or \(2 \times 2 \times 2 \times 2 \times 2 \).

In simplifying an expression like \(\frac{8}{\sqrt[5]{2}}\), find a harmonizing number such that 2 multiplied by it becomes a perfect fifth power. In this instance, it's 16, since

• \[ 2 \times 16 = 32 \]
• which is \(2^5\).

Therefore, multiplying both the top and bottom by \(\sqrt[5]{16}\) ensures the fifth root vanishes from the denominator.