Problem 82 - Elementary and Intermediate Algebra

Question

Rationalize each denominator. All variables represent positive real numbers. $$ \sqrt[3]{\frac{7}{16}} $$

Step-by-Step Solution

Answer

The rationalized form is \( \frac{\sqrt[3]{28}}{4} \).

Step 1: Understand the Expression

The expression given is \( \sqrt[3]{\frac{7}{16}} \), which means finding the cube root of the fraction \( \frac{7}{16} \). We aim to rationalize the denominator, making it a rational number.

Step 2: Simplify the Cube Root Expression

Decompose the cube root of the fraction into the cube root of the numerator and the cube root of the denominator: \( \frac{\sqrt[3]{7}}{\sqrt[3]{16}} \). This step separates the cube root operation from the fraction.

Step 3: Rationalize the Denominator

To rationalize \( \sqrt[3]{16} \), we need to multiply both the numerator and denominator by a suitable expression that will make the denominator a perfect cube. Multiply by \( \sqrt[3]{4} \): \(\frac{\sqrt[3]{7} \cdot \sqrt[3]{4}}{\sqrt[3]{16} \cdot \sqrt[3]{4}} = \frac{\sqrt[3]{28}}{\sqrt[3]{64}} \)This combination makes the denominator \( \sqrt[3]{64} \), which simplifies to a whole number because \( 64 = 4^3 \).

Step 4: Simplify Further

Simplify \( \sqrt[3]{64} \) to 4, making the new expression: \(\frac{\sqrt[3]{28}}{4} \)This form has the rationalized denominator, completing the process.

Step 5: Present the Rationalized Result

The rationalized form of \( \sqrt[3]{\frac{7}{16}} \) is \( \frac{\sqrt[3]{28}}{4} \). Both the numerator and the denominator are in their simplest forms.

Key Concepts

Cube Roots

Cube roots are special because they undo the multiplication of a number by itself twice. It’s a way of finding out which number, when used three times in a multiplication, will give you the original cube. You can liken it to inverse multiplication. The notation for a cube root is usually written as \( \sqrt[3]{x} \).

For example, \( \sqrt[3]{8} = 2 \) since multiplying 2 by itself two more times (\( 2 \times 2 \times 2 \)) results in 8. This is similar to how a square root works, but it involves three instead of two factors.

When working with fractions, you can find the cube root of a fraction by separately taking the cube root of the numerator and the denominator, just like in the expression \( \sqrt[3]{\frac{7}{16}} \). Cube roots can help simplify complex expressions and make them easier to work with, especially in science and engineering where precise values are important.

Rational Numbers

In mathematics, rational numbers are numbers that you can express as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b eq 0 \). This includes numbers like 0.75 or -5, because you can express them as \( \frac{3}{4} \) and \( \frac{-5}{1} \) respectively.

One key property of rational numbers is that their decimal form either terminates or repeats. For instance, the number 0.333... (repeating) is equivalent to \( \frac{1}{3} \), a rational number.

In the context of rationalizing denominators, the goal often is to eliminate any roots or irrational numbers present in the denominator to make it a rational number. This process can simplify calculations and make the expression easier to understand and handle.

Perfect Cubes

Perfect cubes refer to numbers that can be expressed as the cube of an integer. In simple terms, it's a product of an integer multiplied by itself twice, like \( n \times n \times n \). For example, \( 8 = 2^3 \) and \( 27 = 3^3 \) are perfect cubes because they result from the cubing of the integers 2 and 3, respectively.

Identifying a perfect cube can be particularly useful in solving problems involving cube roots, because cube roots of perfect cubes are neat integer numbers. This characteristic is leveraged when rationalizing denominators.

• To make an irrational denominator, such as \( \sqrt[3]{16} \), into a nice integer, you might multiply it by the cube root of another number to turn it into a perfect cube.
• This approach simplifies the calculation greatly, as seen in expressions like \( \frac{\sqrt[3]{28}}{\sqrt[3]{64}} \), where \( 64 \) as a perfect cube breaks down to \( 4^3 \).
• The result, \( \sqrt[3]{64} = 4 \), is a clear integer.

Understanding perfect cubes thus makes tackling cube root problems much more straightforward and enhances arithmetic coherence.