Problem 84 - Intermediate Algebra

Question

Simplify each expression, if possible. All variables represent positive real numbers. $$ 3 \sqrt[3]{27}+12 \sqrt[3]{216} $$

Step-by-Step Solution

Answer

The simplified expression is 81.

Step 1: Simplify each cube root expression

First, let's simplify each of the cube root expressions separately. We'll start with \( \sqrt[3]{27} \). Since \( 27 = 3^3 \), \( \sqrt[3]{27} = 3 \). Next, consider \( \sqrt[3]{216} \). The number 216 can be expressed as \( 6^3 \) because 6 multiplied by itself three times is 216 (i.e., \(6 \times 6 \times 6 = 216\)). Therefore, \( \sqrt[3]{216} = 6 \).

Step 2: Substitute simplified values back into the expression

Now we substitute back the values from the previous step into the original expression. Substitute \( \sqrt[3]{27} \) with 3 and \( \sqrt[3]{216} \) with 6. The expression becomes:\[ 3 \times 3 + 12 \times 6 \]

Step 3: Perform the arithmetic operations

Now, compute the values of the expression. First, calculate the multiplication:\( 3 \times 3 = 9 \) and \( 12 \times 6 = 72 \).Then, add the results together:\( 9 + 72 = 81 \).

Key Concepts

Cube Roots

The concept of cube roots is fundamental when dealing with expressions that involve powers and roots. Essentially, finding the cube root of a number means determining what number, when multiplied by itself three times, results in the original number. For example, to find the cube root of 27, we ask what number satisfies the equation:

• \( x^3 = 27 \)

In this case, \( x = 3 \) because \( 3^3 = 27 \).
Cube roots can be simplified when the number inside the root is a perfect cube, as in our example with 27, which simplifies directly to 3, and 216, which simplifies to 6 because \(6^3 = 216 \).
Simplifying cube roots often involves recognizing perfect cubes and taking advantage of the fact that

• \( \sqrt[3]{a^3} = a \)

when \( a \) is a real number.
This step is crucial before proceeding to arithmetic operations on expressions that include cube roots.

Arithmetic Operations

Arithmetic operations involve basic calculations like addition, subtraction, multiplication, and division. These operations are straightforward, but they follow a specific order known as the Order of Operations, usually remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)).
In our example, after simplifying the cube roots, the expression becomes:

• \( 3 \times 3 + 12 \times 6 \)

Here, we first handle the multiplication:

• \( 3 \times 3 = 9 \)
• \( 12 \times 6 = 72 \)

Next, we add the results:

• \( 9 + 72 = 81 \)

Executing each step sequentially helps ensure accuracy and simplifies the process of working with more complex expressions.

Variables as Positive Reals

When working with algebraic expressions, variables can stand in place of numbers. In the context of the given exercise, these variables represent positive real numbers.
This is important because certain mathematical operations, like finding cube roots, only make sense in the domain of real numbers when dealing with positive values. For instance, the cube root of a positive real number remains a real number.
Understanding that variables in this problem are positive reals helps simplify them correctly and ensures the operations we perform are valid. In situations requiring simplification, treating variables as positive makes it possible to apply traditional arithmetic and root simplification rules without running into undefined or complex numbers, thus maintaining correctness throughout the solution.