Problem 85 - Algebra 2

Question

Rationalize the denominator of each expression. Assume that all variables are positive. \(\frac{\sqrt[3]{x}}{\sqrt[3]{3 y}}\)

Step-by-Step Solution

Answer

\(\frac{\sqrt[3]{9x y^2}}{3y}\)

Step 1: Identify the Root in the Denominator

The given expression is \(\frac{\sqrt[3]{x}}{\sqrt[3]{3 y}}\). The root in the denominator is cube root.

Step 2: Rationalize the Denominator

Rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. As our denominator is a cube root, taking the cube of the denominator will eliminate the root. Hence multiple both the numerator and denominator by \(\sqrt[3]{(3y)^2}\). Now our expression will be \(\frac{\sqrt[3]{x} \cdot \sqrt[3]{(3y)^2}}{\sqrt[3]{3 y} \cdot \sqrt[3]{(3y)^2}}\)

Step 3: Simplify the Expression

Simplify this expression by joining the cube roots in the numerator. The expression then becomes, \(\frac{\sqrt[3]{x \cdot (3y)^2}}{\sqrt[3]{(3 y)^3}}\). As \((3y)^3\) in the denominator simplifies to \(3y\), the rationalized expression can be written as \(\frac{\sqrt[3]{9x y^2}}{3y}\)

Key Concepts

Cube Roots

Cube roots are vital in mathematics as they offer a way to reverse the process of cubing a number. When we deal with cube roots, we are essentially looking for a number that multiplies by itself three times to give the original number. For example, if you take the cube root of 27, you get 3 because

• 3 x 3 x 3 = 27.

In our exercise, we find the cube roots in both the numerator and the denominator. The cube root of an expression like \( \sqrt[3]{x} \) signifies a number which when cubed returns x. It becomes important when rationalizing the denominator, as we must handle these roots carefully to simplify expressions effectively.
Note that unlike square roots, which require a pair, cube roots work with sets of three. This characteristic makes the process of rationalizing denominators with cube roots slightly different.

Mathematical Expressions

Mathematical expressions are combinations of numbers, variables, and operations. They are designed to represent a value or a relationship. In our original problem, \( \frac{\sqrt[3]{x}}{\sqrt[3]{3y}} \) is a mathematical expression comprising variables and cube roots. Understanding how to manipulate these expressions is crucial for solving many mathematics problems.
To work with these expressions, we must follow specific rules and operations. Summing, subtracting, multiplying, or dividing math expressions requires an understanding of order and operations rules, including rationalizing denominators when roots are involved.
When manipulating an expression, it can often be helpful to identify similar terms or factors, use them to your advantage, and remember to apply algebra basics like the distributive property when required. Here, recognizing cube roots helped rationalize the denominator as a part of simplifying the expression.

Simplifying Expressions

Simplifying expressions is about reducing them to a form that is easier to work with while retaining the same value. The original expression, \( \frac{\sqrt[3]{x} \cdot \sqrt[3]{(3y)^2}}{\sqrt[3]{(3y)^3}} \), undergoes simplification to make calculations easier and to present results straightforwardly.
The idea is to reduce any complexity, particularly with denominators, into simpler terms. In our case, rationalizing first involved removing the cube root from the denominator using multiplication by \( \sqrt[3]{(3y)^2} \). This action allowed us to work with simpler expressions like \( \frac{\sqrt[3]{9xy^2}}{3y} \) by canceling out the root in the denominator.

• Focus on eliminating roots when they appear on the bottom part (denominator) of fractions.
• Look for opportunities to combine like terms or factors under a single root when simplifying.
• Once simplified, expressions become much more manageable and easier to interpret or solve in further mathematical contexts.

The simplification process aids in understanding and solving more complex equations by serving up expressions in their most foundational and straightforward form.