Problem 12 - Precalculus: Functions and Graphs

Question

Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt{\left.\sqrt[3]{(} c^{3} d^{6}\right)^{4}}$$

Step-by-Step Solution

Answer

The simplified expression is \(c^{2} d^{4}\).

Step 1: Simplify the Expression Inside the Cube Root

We start by simplifying the expression inside the cube root, which is \( \sqrt[3]{(} c^{3} d^{6})^{4} \). To do this, notice that \((c^{3} d^{6})^{4} = c^{12} d^{24}\). This is because when you raise a power to another power, you multiply the exponents: \((c^{3})^{4} = c^{12}\) and \((d^{6})^{4} = d^{24}\).

Step 2: Simplify Inside the Square Root

After simplifying inside the cube root, we have \(\sqrt{ (c^{12} d^{24})^{1/3} }\). This is the result of taking the cube root of the terms. Using the cube root property, each exponent is divided by 3, giving \(c^{4} d^{8}\) since \(12/3 = 4\) and \(24/3 = 8\).

Step 3: Evaluate the Square Root

Now we have \(\sqrt{c^{4} d^{8}}\). The square root of a term with an even exponent is found by halving the exponent: \(\sqrt{c^{4}} = c^{2}\) and \(\sqrt{d^{8}} = d^{4}\). This simplifies the expression to \(c^{2} d^{4}\).

Step 4: Final Simplification

The final step is to ensure there is no further simplification needed. In this case, \(c^{2} d^{4}\) is already in its simplest form, with no radicals left and no rationalization required, as there is no denominator.

Key Concepts

Cube Roots

A cube root solves an interesting puzzle: what number multiplied by itself three times yields the original number? The cube root operation is denoted by the radical symbol with a small 3, like this: \( \sqrt[3]{x} \). It’s also written as \( x^{1/3} \).
When working with expressions that involve cube roots, especially those within powers, the process may seem challenging initially. Key points to remember when simplifying such expressions are:

• To extract a cube root like \( \sqrt[3]{(} c^{3} d^{6})^{4} \), first simplify inside the cube root if possible.
• Look for cases where the powers or exponents can be broken down by factoring or other operations.

Using these steps systematically allows one not to simplify more straightforwardly but to fully understand the essence of cube roots.

Exponents

Exponents are numbers that reveal how many times a base number is multiplied by itself. In mathematical terms, \( a^b \) means \( a \) is multiplied by itself \( b \) times. Exponents have several handy properties:

• "Product of powers" – \( a^m \cdot a^n = a^{m+n} \)
• "Power of a power" – \((a^m)^n = a^{m \cdot n}\)

In the expression \((c^{3} d^{6})^{4} \), apply the "power of a power" rule to calculate \( c^{12} \) and \( d^{24} \). Understanding these rules makes simplifying complex expressions manageable.

Square Roots

Square roots can be a simpler concept, but they are equally essential in advancing your understanding of expressions. A square root, represented as \( \sqrt{x} \), asks the question: what number squared gives \( x \)?
Simplifying expressions such as \( \sqrt{c^{4} d^{8}} \) involves knowing that taking the square root of an even power means dividing the exponent by two. More specifically:

• \( \sqrt{c^{4}} = c^{2} \) because \( c^{2} \cdot c^{2} = c^{4} \)
• \( \sqrt{d^{8}} = d^{4} \)

This technique lets you unravel expressions into their simplest forms, which is tremendously beneficial in mathematics.

Rationalizing Denominators

Rationalizing denominators may sound complex, but it is a technique that ensures expressions do not have irrational numbers (like square roots) in the denominator. While the current exercise does not involve denominators, understanding this concept is valuable. Rationalizing involves multiplying both the numerator and the denominator by a term that will eliminate the radical in the denominator. For example:

• If the denominator is \( \sqrt{a} \), multiply by \( \sqrt{a} \).
• For a complex term \( \sqrt{a} + b \), multiply by its conjugate, \( \sqrt{a} - b \).

This results in a simpler, cleaner expression that is fully rational. Familiarity with this process is vital as it frequently appears in algebra and calculus.