Problem 63 - College Algebra

Question

Simplify each expression. $$\sqrt[3]{128 z^{3}}-\sqrt[3]{-16 z^{3}}$$

Step-by-Step Solution

Answer

6z \cdot \sqrt[3]{2}

Step 1: Simplify Each Cube Root Separately

First, simplify the expression inside each cube root separately. For \( \sqrt[3]{128z^3} \), factor 128 into its prime components to get \( \sqrt[3]{2^7 z^3} \). Recognize that 7 can be split into 6 (which is a multiple of 3) and 1, so you rewrite it as \( \sqrt[3]{(2^6 \cdot 2^1) z^3} \). Similarly, for \( \sqrt[3]{-16z^3} \), factor -16 into its prime components as \( \sqrt[3]{-(2^4)z^3} \).

Step 2: Extract Perfect Cubes

Extract the cube roots of the perfect cubes in both expressions. \( \sqrt[3]{(2^6 \cdot 2^1) z^3} = 2^2 \cdot 2 \cdot z = 4z \cdot \sqrt[3]{2} \), since \( (2^6) \) is a perfect cube whose cube root is \( 2^2 \), and the cube root of \( z^3 \) is \( z \). Similarly, for \( \sqrt[3]{-(2^4)z^3} \), we get \( \sqrt[3]{-(2^3 \cdot 2^1) z^3} = -2z \cdot \sqrt[3]{2}\).

Step 3: Simplify the Subtraction

Subtract the two simplified expressions: \( (4z \cdot \sqrt[3]{2}) - (-2z \cdot \sqrt[3]{2}) \). This simplifies to \( 4z \cdot \sqrt[3]{2} + 2z \cdot \sqrt[3]{2} \).

Step 4: Combine Like Terms

Combine the like terms involving \( z \cdot \sqrt[3]{2} \). The result is \( (4z + 2z) \cdot \sqrt[3]{2} = 6z \cdot \sqrt[3]{2} \).

Key Concepts

Cube Roots

The cube root of a number is a value that, when multiplied by itself twice, gives the original number. Essentially, if \(x^3 = a\), then \(x = \sqrt[3]{a}\). This concept is often used to simplify expressions in algebra. Unlike square roots, cube roots can result in both positive and negative values since multiplying a negative number by itself three times results in a negative number.
For example, \(\sqrt[3]{8} = 2\) because \(2 \times 2 \times 2 = 8\). Similarly, \(\sqrt[3]{-27} = -3\) because \(-3 \times -3 \times -3 = -27\). When simplifying expressions involving cube roots, it helps to recognize perfect cubes, which are numbers like 8, 27, and 64 that can be cubed to give a whole number.

Prime Factorization

Prime factorization is the process of breaking down a number into its basic building blocks, which are prime numbers. A prime number is one that can only be divided by 1 and itself, such as 2, 3, 5, 7, and so on. This method is particularly helpful in finding cube roots of numbers by identifying perfect cubes within large numbers.
For example, let's take the number 128. We can express it as a product of prime factors like \(128 = 2^7\). By doing this, we can easily identify parts of the number that form perfect cubes. In cube root simplification, we look for groups of three identical factors, such as \(2^3\), to simplify the expression further. Understanding how to break down a number into its prime factors is a key skill in algebra for simplifying not only cube roots but other algebraic expressions as well.

Simplifying Expressions

Simplifying expressions in algebra involves reducing them to their simplest form. When we work with cube roots, we first identify factors that can be simplified, such as perfect cubes. This makes the expression easier to handle and understand.
For example, simplifying \(\sqrt[3]{128z^3}\) involves using the prime factorization of 128, which we found to be \(2^7\). We then separate the perfect cube portion \(2^6\) from \(2^1\) and calculate as follows: \(\sqrt[3]{2^6 \cdot 2^1 \cdot z^3} = 4z \cdot \sqrt[3]{2}\). Through each step, we focus on simplifying the parts that can be easily dealt with, which often involves pulling out whole numbers from cube roots and handling like terms effectively.

Perfect Cubes

A perfect cube is a number that can be expressed as \(x^3\), where \(x\) is an integer. Common perfect cubes include 1 (1x1x1), 8 (2x2x2), 27 (3x3x3), and so forth. Recognizing perfect cubes helps in solving problems where cube roots are involved.
For the expressions in our example, recognizing \(2^6\) as a perfect cube is crucial. Since \(2^2 \cdot 2^2 \cdot 2^2 = 2^6\), its cube root simplifies to \(2^2 = 4\). Similarly, \(-(2^3)\) is a perfect cube because its cube root is \(-2\), thanks to \((-2)\times(-2)\times(-2) = -8\). Identifying and extracting perfect cubes speeds up the simplification process, especially in composite algebraic expressions.