Problem 64 - Precalculus: Functions and Graphs

Question

Rationalize the denominator. $$\frac{1}{\sqrt[3]{x}+\sqrt[3]{y}}$$ (Hint: Multiply numerator and denominator $$\text { by } \sqrt[3]{a^{2}}+\sqrt[3]{a b}+\sqrt[3]{b^{2}}$$

Step-by-Step Solution

Answer

Rationalized form: \(\frac{\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2}}{x + y}\).

Step 1: Identify the Expression

We need to rationalize the denominator of the expression \(\frac{1}{\sqrt[3]{x} + \sqrt[3]{y}}\). Currently, the denominator is not in a rational form because it contains cube roots.

Step 2: Recognize Conjugates

To rationalize the denominator, we recognize the form \(a + b\) and the need to multiply it by its corresponding conjugate \(a^2 - ab + b^2\) to eliminate cube roots. For the expression \(\sqrt[3]{x} + \sqrt[3]{y}\), we consider \(a = \sqrt[3]{x}\) and \(b = \sqrt[3]{y}\).

Step 3: Formulate the Conjugate

Determine the conjugate for \(\sqrt[3]{x} + \sqrt[3]{y}\), which is \(\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2}\).

Step 4: Multiply by the Conjugate

Multiply both the numerator and denominator by the conjugate. The new expression becomes:\[\frac{1 \cdot (\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2})}{(\sqrt[3]{x} + \sqrt[3]{y})(\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2})}\]

Step 5: Simplify the Denominator

Multiply the denominator using the formula \((a + b)(a^2 - ab + b^2) = a^3 + b^3\). Here, \(a = \sqrt[3]{x}\) and \(b = \sqrt[3]{y}\), so the denominator simplifies to \(x + y\).

Step 6: Write the Final Expression

The expression is now simplified to: \[\frac{\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2}}{x + y}\], which successfully rationalizes the denominator.

Key Concepts

Cube Roots

Cube roots are similar to square roots, but instead of finding a number that multiplies by itself twice to form a given number, you are looking for a number that multiplies by itself three times. For example, the cube root of 8 is 2 because 2 multiplied by itself three times (2 × 2 × 2) equals 8. In mathematics, cube roots are expressed with a radical sign and a small three above it, such as \(\sqrt[3]{x}\). Cube roots can often make calculations complicated, especially when they appear in the denominator of a fraction. This complexity is precisely why rationalizing the denominator is a helpful technique. By getting rid of cube roots in the denominator, arithmetic with these expressions becomes much more manageable.

Conjugates in Algebra

Conjugates are a vital tool in algebra, especially for simplifying expressions involving roots. When dealing with square roots, the conjugate of an expression \(a + b\) is \(a - b\). However, when cube roots are involved, the concept of a conjugate is slightly different. For an expression like \(a + b\) where both are cube roots, its conjugate is typically expressed as \(a^2 - ab + b^2\). This works because multiplying \(a + b\) by its conjugate \(a^2 - ab + b^2\) eliminates the cube roots in favor of a simpler expression. This concept is crucial in rationalizing denominators involving cube roots as it effectively simplifies the expression to something you can work with more easily, such as another rational expression.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. A key part of working with rational expressions is ensuring they are in their simplest form. This often includes rationalizing the denominator, which means manipulating the expression so that the denominator contains no radicals or roots. This process is especially important when the denominator includes cube roots because it converts the expression into a format that is often easier to understand and use in further mathematical operations. By removing roots from the denominator, you ensure that you can perform arithmetic on these expressions more straightforwardly, aiding in clearer and more efficient problem-solving.