Problem 53 - A Graphical Approach to College Algebra

Question

If possible, simplify each radical expression. Assume that all variables represent positive real numbers. $$\frac{\sqrt[3]{m n} \cdot \sqrt[3]{m^{2}}}{\sqrt[3]{n^{2}}}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{m}{\sqrt[3]{n}}\).

Step 1: Multiply the Radicals in the Numerator

We start by combining the cube roots in the numerator. The expression is \(\sqrt[3]{m n} \cdot \sqrt[3]{m^{2}}\). According to the property of radicals \(\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a \cdot b}\), they can be combined into a single cube root: \(\sqrt[3]{(m n) \cdot m^{2}} = \sqrt[3]{m^{3}n}\).

Step 2: Simplify the Cube Root in the Numerator

Next, simplify \(\sqrt[3]{m^{3}n}\). Since \(m^{3}\) can be simplified to \(m\) (since the cube root and the cube cancel each other), the expression becomes \(m\sqrt[3]{n}\).

Step 3: Simplify the Entire Expression

Now, rewrite the entire expression, replacing the numerator with the simplified version: \(\frac{m\sqrt[3]{n}}{\sqrt[3]{n^{2}}}\).

Step 4: Simplify the Division of Radical Expressions

Use the property of radicals that states \(\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}\). Apply this to the expression: \(m\sqrt[3]{\frac{n}{n^{2}}} = m\sqrt[3]{n^{1-2}} = m\sqrt[3]{n^{-1}}\).

Step 5: Express the Simplified Radical as a Fraction

Since \(\sqrt[3]{n^{-1}} = \frac{1}{\sqrt[3]{n}}\), rewrite the expression as \(\frac{m}{\sqrt[3]{n}}\).

Step 6: Final Simplification

This is the simplest form of the expression \(\frac{m}{\sqrt[3]{n}}\).

Key Concepts

Simplification

When dealing with radical expressions, simplification is about making them easier to understand or work with. It often involves reducing expressions to their simplest form. In the context of cube roots, we aim to combine and simplify the components step by step.

• Identify and group similar terms or operations in the expression.
• Apply specific algebraic properties, like the product property of radicals, to combine terms.
• Simplify the expression by canceling out common factors.

Let's look at the expression from the problem: \[\frac{\sqrt[3]{m n} \cdot \sqrt[3]{m^{2}}}{\sqrt[3]{n^{2}}}\]We start by using the property of radicals to combine the cube roots in the numerator. This reduces the multiple radicals into a single radical, making the expression easier to handle. By further simplifying, we are able to reduce the expression to: \[\frac{m}{\sqrt[3]{n}}\] which is its simplest form.

Cube Roots

Cube roots are a type of root similar to square roots but instead of finding what number multiplied by itself gives the square, we find what number multiplied by itself twice, results in the original number. This concept is fundamental when simplifying radical expressions with cubes, just like in our example.

• The cube root of a number \( a \) is written as \( \sqrt[3]{a} \).
• For instance, the cube root of \( 27 \) is \( 3 \), because \( 3 \times 3 \times 3 = 27 \).
• Cube roots of expressions can be handled using properties like \( \sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab} \), which simplifies multiplication under the radical.

In the problem, we used cube root simplification when combining terms: \[\sqrt[3]{m n} \cdot \sqrt[3]{m^{2}} = \sqrt[3]{m^3n} \]which then simplifies further because the *cube* of \( m \) "cancels" with the cube root.

Algebraic Properties

Understanding algebraic properties is crucial when simplifying radical expressions. These properties allow you to manipulate and simplify expressions systematically.

• Product Property: \( \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a \cdot b} \) allows for the combination of terms under a common radical.
• Quotient Property: \( \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}} \), useful for simplifying fractions under a radical.

These principles were applied in the step-by-step solution to our exercise. For example, by using the quotient property, the expression \[\frac{m\sqrt[3]{n}}{\sqrt[3]{n^{2}}}\] was converted into \[ m\sqrt[3]{\frac{n}{n^{2}}} \] and further simplified by adjusting exponents, showcasing the power of understanding and applying these foundational concepts.