Problem 27 - Algebra and Trigonometry

Question

Simplify the expression. \(\sqrt[3]{108}-\sqrt[3]{32}\)

Step-by-Step Solution

Answer

The expression simplifies to \(\sqrt[3]{4}\).

Step 1: Prime Factorization of 108

First, we need to find the prime factorization of 108. We do this by dividing by the smallest prime numbers: 108 is even, so divide by 2 to get 54. Continuing, 54 divided by 2 is 27. Then, 27 divided by 3 is 9, and finally, 9 divided by 3 is 3. So, the prime factorization of 108 is \(2^2 \times 3^3\).

Step 2: Apply Cube Root to 108

To simplify \(\sqrt[3]{108}\), use the prime factorization: \(\sqrt[3]{2^2 \times 3^3} = \sqrt[3]{2^2} \times \sqrt[3]{3^3}\). Since \(\sqrt[3]{3^3} = 3\), we get \(3 \times \sqrt[3]{2^2} = 3 \times \sqrt[3]{4}\).

Step 3: Prime Factorization of 32

Now, find the prime factorization of 32. 32 is even, so divide by 2 to get 16, then continue dividing by 2: 16 \(\div\) 2 = 8, 8 \(\div\) 2 = 4, 4 \(\div\) 2 = 2. Thus, the prime factorization is \(2^5\).

Step 4: Apply Cube Root to 32

Simplify \(\sqrt[3]{32}\) using its prime factorization: \(\sqrt[3]{2^5} = \sqrt[3]{2^3} \times \sqrt[3]{2^2}\). Since \(\sqrt[3]{2^3} = 2\), we have \(2 \times \sqrt[3]{4}\).

Step 5: Simplify the Entire Expression

Combine the simplified cube roots from Steps 2 and 4: \(3 \times \sqrt[3]{4} - 2 \times \sqrt[3]{4}\). Factor out \(\sqrt[3]{4}\) to get \((3 - 2) \sqrt[3]{4} = 1 \times \sqrt[3]{4}\). Therefore, the expression simplifies to \(\sqrt[3]{4}\).

Key Concepts

Prime Factorization

Prime factorization involves breaking a number down into its basic building blocks, which are the prime numbers. A prime number is any number greater than 1 that cannot be divided by any number other than itself and 1. The prime factorization of a number is the expression of that number as a product of its prime factors. This method is crucial for multiple mathematical operations, such as finding cube roots, as it allows for simplification of complex expressions.

For instance, the number 108 can be expressed using its prime factors as follows:

• 108 is divisible by 2, giving us 54.
• 54 is divisible by 2 again, resulting in 27.
• Next, 27 is divisible by 3, going down to 9.
• Finally, dividing 9 by 3 gives us 3.

So, the prime factorization of 108 is \(2^2 \times 3^3\). This step-by-step breakdown helps us simplify expressions and solve problems involving cube roots more effectively. Similarly, when we factorize 32, the breakdown is:

• 32 \div 2 = 16
• 16 \div 2 = 8
• 8 \div 2 = 4
• 4 \div 2 = 2

This means the prime factorization of 32 is \(2^5\). By employing prime factorization, we can simplify expressions involving cube roots by transforming them into a more workable form.

Simplifying Expressions

Simplifying expressions allows us to work with numbers or equations in a more manageable way. This often involves reducing numbers to their simplest form or combining like terms. When it comes to cube roots, simplifying involves breaking down numbers using their prime factorization and applying the cube root operation directly to these simpler forms.

Consider the expression \(\sqrt[3]{108}\). Using the prime factorization \(2^2 \times 3^3\), we take the cube root of each factor:

• \(\sqrt[3]{2^2}\)
• \(\sqrt[3]{3^3} = 3\)

This gives us \(3 \times \sqrt[3]{4}\), a simpler form to work with. Similarly, for \(\sqrt[3]{32}\), we use the factor \(2^5\):

• \(\sqrt[3]{2^3} = 2\)
• \(\sqrt[3]{2^2} = \sqrt[3]{4}\)

Thus, \(\sqrt[3]{32} = 2 \times \sqrt[3]{4}\). These simplified expressions make it easier to combine them during subtraction, leading to a final simplified expression.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. In the context of cube roots, algebraic expressions can be used to represent these roots in a simplified form. Understanding how to manipulate these expressions is fundamental to solving problems effectively.

When we have an expression like \(\sqrt[3]{108} - \sqrt[3]{32}\), we're dealing with the subtraction of algebraic expressions. Each part of this expression has been simplified using prime factorization, resulting in:

• \(3 \times \sqrt[3]{4}\) from \(\sqrt[3]{108}\)
• \(2 \times \sqrt[3]{4}\) from \(\sqrt[3]{32}\)

By using algebra, we can now see that both parts share \(\sqrt[3]{4}\) as a common factor. We factor out \(\sqrt[3]{4}\) and perform the subtraction: \((3 - 2) \times \sqrt[3]{4} = 1 \times \sqrt[3]{4}\).

Understanding algebraic expressions allows us to manipulate and simplify complex mathematical ideas by breaking them down into manageable parts. This skill is particularly useful in making seemingly difficult problems more approachable.