Chapter 0: Problem 20
Simplify each expression. Assume that all variables are positive when they appear. $$ \sqrt[3]{54}$$
Short Answer
Expert verified
3 \sqrt[3]{2}.
Step by step solution
01
Find the Prime Factorization of 54
First, determine the prime factors of 54. 54 can be divided by 2 (since it is even), so we get 54 = 2 * 27. Next, factor 27, which is 27 = 3^3. Therefore, 54 = 2 * 3^3.
02
Simplify the Expression Using Cube Root Properties
Apply the cube root to the prime factorization of 54, using the property \(\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}\). So, we get \(\sqrt[3]{54} = \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3}\).
03
Evaluate the Cube Roots
Evaluate \(\sqrt[3]{3^3}\). Since the cube root of a cubed number is the number itself, \(\sqrt[3]{3^3} = 3\). Therefore, \(\sqrt[3]{54} = \sqrt[3]{2} \times 3\).
04
Combine the Simplified Parts
Combine the simplified parts to find the final answer: \(\sqrt[3]{54} = 3 \sqrt[3]{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Prime Factorization
Prime factorization is the process of expressing a number as the product of its prime factors. Prime factors are numbers that are only divisible by 1 and themselves. For example, to find the prime factorization of 54, we start by dividing 54 by the smallest prime number 2 (since 54 is even):
54 ÷ 2 = 27
Next, we factor 27. The prime factors of 27 are all 3's (since 3 is a prime number):
27 = 3 × 3 × 3 or 3^3.
Combining these results, we get the prime factorization of 54 as:
54 = 2 * 3^3.
54 ÷ 2 = 27
Next, we factor 27. The prime factors of 27 are all 3's (since 3 is a prime number):
27 = 3 × 3 × 3 or 3^3.
Combining these results, we get the prime factorization of 54 as:
54 = 2 * 3^3.
Cube Root Properties
Understanding cube root properties is essential for simplifying cube roots. A cube root is written as \( \sqrt[3]{} \) and is the number that, when multiplied by itself three times, gives the original number. Some key properties include:
- \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \)
- \( \sqrt[3]{a^3} = a \)
Simplifying Radicals
Simplifying radicals involves restructuring the expression to its simplest form. When simplifying cube roots, we use the prime factorization and cube root properties:
For example, simplify \( \sqrt[3]{54} \):
Given the prime factorization, 54 = 2 * 3^3, apply the cube root to both parts:
\( \sqrt[3]{54} = \sqrt[3]{2 \times 3^3} \).
Using \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \), we get:
\( \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3} \).
Since \( \sqrt[3]{3^3} = 3 \), it simplifies to:
\( \sqrt[3]{2} \times 3 \).
Thus, \( \sqrt[3]{54} = 3\sqrt[3]{2} \).
For example, simplify \( \sqrt[3]{54} \):
Given the prime factorization, 54 = 2 * 3^3, apply the cube root to both parts:
\( \sqrt[3]{54} = \sqrt[3]{2 \times 3^3} \).
Using \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \), we get:
\( \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3} \).
Since \( \sqrt[3]{3^3} = 3 \), it simplifies to:
\( \sqrt[3]{2} \times 3 \).
Thus, \( \sqrt[3]{54} = 3\sqrt[3]{2} \).
Evaluating Cube Roots
Evaluating cube roots involves finding the number that, when cubed, gives the original number. Here’s how you can evaluate \( \sqrt[3]{54} \):
After simplifying, we broke it down to \( \sqrt[3]{2} \times \sqrt[3]{3^3} \). Since the cube root of a cubed number is the number itself:
\( \sqrt[3]{3^3} = 3 \).
Thus, it's easier to evaluate real numbers directly from their prime factors:
So, \( \sqrt[3]{54} = \sqrt[3]{2} \times 3 \).
This simplifies to:
\( 3\sqrt[3]{2} \).
By breaking it down into simpler steps, you can handle even complex cube roots efficiently.
After simplifying, we broke it down to \( \sqrt[3]{2} \times \sqrt[3]{3^3} \). Since the cube root of a cubed number is the number itself:
\( \sqrt[3]{3^3} = 3 \).
Thus, it's easier to evaluate real numbers directly from their prime factors:
So, \( \sqrt[3]{54} = \sqrt[3]{2} \times 3 \).
This simplifies to:
\( 3\sqrt[3]{2} \).
By breaking it down into simpler steps, you can handle even complex cube roots efficiently.