Problem 27 - Technical Mathematics with Calculus

Question

Multiplication of Radicals. Multiply and simplify. $$2 \sqrt[3]{24} \text{ by } \sqrt[9]{\frac{8}{27}}$$

Step-by-Step Solution

Answer

\(\frac{8}{3} \sqrt[3]{3}\)

Step 1: Express the Radicands in Prime Factorization

Break down the numbers inside the radicals into their prime factors. For 24, the prime factorization is 2 * 2 * 2 * 3. For 8, the prime factorization is 2 * 2 * 2, and for 27 it is 3 * 3 * 3.

Step 2: Rewrite the Radicals with Prime Factors

Replace the numbers inside the radicals with their prime factorizations. Thus, we have \(2 \sqrt[3]{2 \cdot 2 \cdot 2 \cdot 3}\) and \(\sqrt[9]{\frac{2 \cdot 2 \cdot 2}{3 \cdot 3 \cdot 3}}\).

Step 3: Simplify the Cube Root

Since we have a cube root, we look for groups of three identical factors to take out of the root. For \(\sqrt[3]{2 \cdot 2 \cdot 2 \cdot 3}\), we can take out a group of three 2s and it becomes \(2 \cdot 2 \sqrt[3]{3}\).

Step 4: Simplify the Ninth Root

Similarly, for the ninth root, we have three sets of three 2s and three sets of three 3s within the fraction. This simplifies to \(\sqrt[9]{\frac{2 \cdot 2 \cdot 2}{3 \cdot 3 \cdot 3}} = \sqrt[9]{\frac{2^3}{3^3}} = \frac{2}{3}\).

Step 5: Execute the Multiplication

Now we multiply \(2 \cdot 2 \sqrt[3]{3}\) by \(\frac{2}{3}\). It simplifies to \(\frac{4 \cdot 2}{3} \sqrt[3]{3}\), which further simplifies to \(\frac{8}{3} \sqrt[3]{3}\).

Step 6: Express the Final Answer

Combine the simplified terms to express the final simplified form of the multiplication, which is \(\frac{8}{3} \sqrt[3]{3}\).

Key Concepts

Prime Factorization

Prime factorization is a process used to break down a composite number into its constituent prime factors, which are the building blocks of all natural numbers. Prime numbers are numbers that have exactly two distinct positive divisors: one and itself. In the multiplication of radicals, prime factorization allows us to see the structure within a radical, which can be crucial when simplifying expressions.

For instance, when faced with a problem involving the cube root of 24, breaking it down into prime factors (2 × 2 × 2 × 3) gives a clearer view of which numbers can be pulled out of the radical when simplifying. These prime factors are the 'atomic' parts which we can't break down any further in the number system and are extremely useful in many areas of mathematics, especially in the context of simplifying radicals.

Simplify Cube Root

The cube root of a number asks the question: 'What number, when multiplied by itself three times, gives the original number?' Simplifying a cube root involves identifying groups of three identical factors within the radicand (the number under the radical symbol) and pulling one of each group out of the radical.

To illustrate, when simplifying the cube root of 24, we first express 24 as its prime factors (2 × 2 × 2 × 3). We find that there's a group of three 2's, which we can take out of the cube root, turning \(\sqrt[3]{2 \cdot 2 \cdot 2 \cdot 3}\) into \(2 \cdot \sqrt[3]{3}\). This process reduces the complexity of the expression, making it easier to handle in further calculations or multiplications.

Simplify Ninth Root

Simplifying a ninth root follows a concept similar to simplifying a cube root, but instead of looking for groups of three, we're looking for groups of nine. The ninth root of a number answers this question: 'What number when raised to the power of nine results in the original number?'

Let's consider the ninth root of \(\frac{8}{27}\). By performing prime factorization, we write 8 as (2 × 2 × 2) and 27 as (3 × 3 × 3). Our ninth root becomes \(\sqrt[9]{\frac{2^3}{3^3}}\), suggesting each group of three factors represents a single number once taken out of the root. Thus, the ninth root of \(\frac{8}{27}\) simplifies to \(\frac{2}{3}\). This showcases the power of the prime factorization technique in simplifying seemingly complex radical expressions.