Problem 92 - Intermediate Algebra

Question

Given \(\frac{\sqrt[3]{5 y}}{\sqrt[3]{4}},\) rationalize the denominator by following parts (a) and (b). a. Multiply the numerator and denominator by \(\sqrt[3]{16}\) b. Multiply the numerator and denominator by \(\sqrt[3]{2}\) c. What can you conclude from parts (a) and (b)?

Step-by-Step Solution

Answer

Both methods yield rational denominators, 4 and 2, with different numerators.

Step 1: Understand the Problem

We are given the expression \(\frac{\sqrt[3]{5y}}{\sqrt[3]{4}}\) and asked to rationalize its denominator using two different multiplication methods, then draw a conclusion from these approaches.

Step 1: Multiply by Cube Root of 16

For part (a), multiply both the numerator and the denominator by \(\sqrt[3]{16}\) to rationalize the denominator:\[ \frac{\sqrt[3]{5y}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{16}}{\sqrt[3]{16}} = \frac{\sqrt[3]{80y}}{\sqrt[3]{64}} \] Evaluating, \(\sqrt[3]{64}\) simplifies to 4, thus the expression becomes \( \frac{\sqrt[3]{80y}}{4} \).

Step 2: Multiply by Cube Root of 2

For part (b), multiply both the numerator and the denominator by \(\sqrt[3]{2}\) to rationalize the denominator:\[ \frac{\sqrt[3]{5y}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{\sqrt[3]{10y}}{\sqrt[3]{8}} \] Evaluating, \(\sqrt[3]{8}\) simplifies to 2, thus the expression becomes \( \frac{\sqrt[3]{10y}}{2} \).

Step 3: Conclusion

From parts (a) and (b), we can conclude that rationalizing the denominator using different cube roots results in different numerators, but both expressions simplify to have integer denominators, specifically 4 and 2, respectively.

Key Concepts

Cube Roots

A cube root is the number that, when multiplied by itself twice, results in a given number. For example, the cube root of 8 is 2 because

• 2 * 2 * 2 = 8.

Cube roots are similar to square roots, but they involve three factors instead of two. This operation is particularly useful in algebraic manipulations where certain expressions with exponents need simplification. In the given problem, we have cube roots in both the numerator and the denominator.
The ultimate aim here is to simplify the denominator to a whole number by manipulating cube roots strategically. We need to multiply both the numerator and the denominator by another cube root that, when multiplied with the current denominator, results in a perfect cube. This simplifies the expression to have an integer under the cube root in the denominator, which can further simplify to a whole number.

Expression Simplification

Expression simplification involves the process of making an expression easier to understand or manipulate while retaining its original value. In algebra, expressions can often include variables, exponents, and roots, as seen in the current problem. The goal is to remove any unnecessary complexity, especially those in the denominators, since they can make calculations cumbersome.
In our example, the expression is

• \( \frac{\sqrt[3]{5y}}{\sqrt[3]{4}} \)

By using cube root multiplication, we can rationalize the denominator. This is achieved through two methods:

• Multiplying by \( \sqrt[3]{16} \), yielding \( \frac{\sqrt[3]{80y}}{4} \).
• Multiplying by \( \sqrt[3]{2} \), resulting in \( \frac{\sqrt[3]{10y}}{2} \).

Both methods successfully simplify the expression to a point where the denominators are integers, making further arithmetic easier.

Algebraic Manipulation

Algebraic manipulation refers to using standard algebraic techniques to rearrange or simplify expressions and equations. This process can involve distributing factors, combining like terms, or rationalizing denominators, as we see in the current exercise.
Rationalizing a denominator means modifying the expression so that the denominator becomes a rational number. In our problem concerning the expression \( \frac{\sqrt[3]{5y}}{\sqrt[3]{4}} \):

• We multiply the expression by \( \sqrt[3]{16} \) in part (a) to simplify the denominator to 4 because \( \sqrt[3]{4} \times \sqrt[3]{16} = \sqrt[3]{64} = 4 \).
• In part (b), multiplying by \( \sqrt[3]{2} \) simplifies the denominator to \( 2 \) because \( \sqrt[3]{4} \times \sqrt[3]{2} = \sqrt[3]{8} = 2 \).

Both approaches highlight different paths to get to a similar end: a simplified, integer-based denominator. Hence, showing the versatility of algebraic manipulation.