Problem 29 - Algebra 2 and Trigonometry

Question

In \(3-38\) write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero. $$ \sqrt[3]{54}+\sqrt[3]{128} $$

Step-by-Step Solution

Answer

The simplest form is \( 7\sqrt[3]{2} \).

Step 1: Simplify the radical expression \( \sqrt[3]{54} \)

Begin by factoring 54 into its prime factors: 54 = 2 × 3 × 3 × 3. Rewrite \( \sqrt[3]{54} \) using these factors: \( \sqrt[3]{2 \times 3^3} \). Simplifying, \( \sqrt[3]{54} = \sqrt[3]{3^3 \times 2} = 3\sqrt[3]{2} \).

Step 2: Simplify the radical expression \( \sqrt[3]{128} \)

Break 128 into its prime factors: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^7. Rewrite \( \sqrt[3]{128} \) as \( \sqrt[3]{2^7} \). Simplifying, \( 2^7 = 2^6 \times 2 = (2^2)^3 \times 2 = 2^2 \times \sqrt[3]{2} \). So, \( \sqrt[3]{128} = 4\sqrt[3]{2} \).

Step 3: Combine the simplified expressions

Now combine the simplified terms from Step 1 and Step 2: \( 3\sqrt[3]{2} + 4\sqrt[3]{2} = (3 + 4)\sqrt[3]{2} = 7\sqrt[3]{2} \).

Step 4: Final Step: Write the expression in simplest form

The expression \( \sqrt[3]{54} + \sqrt[3]{128} \) simplifies to \( 7\sqrt[3]{2} \).

Key Concepts

Prime Factorization

Prime factorization is a method used to express a number as the product of its basic building blocks, known as prime numbers. Prime numbers are numbers greater than 1 that have no other divisors except for 1 and themselves. To perform prime factorization, you keep dividing the number by the smallest prime numbers (like 2, 3, 5, etc.) until what you have left is a prime number itself. This method is exceptionally useful when dealing with radical expressions.

In the example given, we performed prime factorization on 54, which resulted in:

• 54 = 2 × 3 × 3 × 3 = 2 × 3^3

Similarly, for 128, the prime factorization was:

• 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^7

After finding the prime factors, these numbers can be more easily simplified within a radical expression which makes solving problems involving cubes or higher powers more manageable.

Cubic Roots

Cubic roots are used when you want to find a number which, when multiplied by itself three times, gives the original number. It is often represented with the cube root symbol \( \sqrt[3]{} \). Finding cube roots can be tricky but becomes simpler once you break the number into its prime factors, especially for perfect cubes.

In simplifying expressions like \( \sqrt[3]{54} \) and \( \sqrt[3]{128} \), once we performed the prime factorization, it allowed us to rewrite the expression in terms of their cubic roots. For example, once 54 is factored as \( 2 \times 3^3 \), we recognize \( 3^3 \) as a perfect cube. This means \( \sqrt[3]{3^3} = 3 \), easily giving us part of our simplification of the cubic root expression as \( 3\sqrt[3]{2} \).

For 128, we use the fact that \( 128 = 2^7 = (2^2)^3 \times 2 \), allowing us to simplify \( \sqrt[3]{128} \) to \( 4\sqrt[3]{2} \) as we extract \( (2^2)^3 \) from under the cube root. Working with cubic roots often involves recognizing these smaller perfect cubes within the number.

Combining Like Terms

Combining like terms is a fundamental practice in algebra which involves adding or subtracting terms in an expression that have the same variable component. This is an important step to simplify expressions and solve equations efficiently.

In the context of the given exercise, we are combining terms that have the same cubic root component of \( \sqrt[3]{2} \). After we simplified:

• \( 3\sqrt[3]{2} \) from \( \sqrt[3]{54} \)
• \( 4\sqrt[3]{2} \) from \( \sqrt[3]{128} \)

We combine the like terms by adding their coefficients, which in this case are 3 and 4, respectively. This yields \( (3 + 4)\sqrt[3]{2} = 7\sqrt[3]{2} \).

Combining like terms reduces the complexity of the expression and is a crucial step in providing a final, simplest form of the expression. This technique is not only useful for radicals but is applied broadly across algebraic expressions to simplify and solve them.