Problem 74
Question
Simplify each cube root. See Example 6. $$ \sqrt[3]{-27} $$
Step-by-Step Solution
Verified Answer
The simplified form of \(
\sqrt[3]{-27}\) is \(-3\).
1Step 1: Identify the Cube Root Problem
The cube root expression given is \(\sqrt[3]{-27}\). A cube root asks us to find a number that, when multiplied by itself three times, gives the original number \(-27\).
2Step 2: Determine the Cube Root of Positive 27
First, determine the cube root of \(27\). We know that \(3 \times 3 \times 3 = 27\). Thus, the cube root of \(27\) is \(3\), or \(\sqrt[3]{27} = 3\).
3Step 3: Apply Cube Root to Negative Number
Since we are dealing with \(-27\), recall that the cube of a negative number is also negative. Therefore, \(-3 \times -3 \times -3 = -27\). Hence, the cube root of \(-27\) is \(-3\).
4Step 4: Write Final Simplified Form
Conclude by stating the simplified form of the original expression: \(\sqrt[3]{-27} = -3\).
Key Concepts
Understanding Cube RootsWorking with Negative NumbersSimplifying Radicals Further
Understanding Cube Roots
When dealing with cube roots, it's essential to understand what they represent. Simply put, a cube root of a number is a value that, when multiplied by itself three times, will result in the original number. For example, consider the number 8. The cube root of 8 is 2 because when you multiply 2 three times, you get 8:
- \[2 \times 2 \times 2 = 8\]
Working with Negative Numbers
Cube roots can apply to both positive and negative numbers. This is because multiplying a negative number by itself three times still results in a negative number, due to the properties of arithmetic operations.
For instance, let's explore the cube root of \(\ -27 \). When looking for what multiplied by itself three times gives \(\ -27 \), we consider \(\ -3 \):
Thus, the concept of cube roots extends neatly to negative numbers without requiring any complex numbers. This highlights an important difference from square roots, which do not yield real results for negative numbers because squares are always positive.
For instance, let's explore the cube root of \(\ -27 \). When looking for what multiplied by itself three times gives \(\ -27 \), we consider \(\ -3 \):
- \[-3 \times -3 \times -3 = -27\]
Thus, the concept of cube roots extends neatly to negative numbers without requiring any complex numbers. This highlights an important difference from square roots, which do not yield real results for negative numbers because squares are always positive.
Simplifying Radicals Further
Simplifying radicals, especially cube roots, involves identifying the prime factors or perfect cubes within a number. By doing so, we can find more straightforward representations of these mathematical expressions.
The key is to determine whether a number is a perfect cube or can be broken into smaller perfect cubes. If a number like 64 can be expressed as a cube, i.e., \((4 \times 4 \times 4)\), it simplifies directly to 4 without further need for estimation or calculation.
The key is to determine whether a number is a perfect cube or can be broken into smaller perfect cubes. If a number like 64 can be expressed as a cube, i.e., \((4 \times 4 \times 4)\), it simplifies directly to 4 without further need for estimation or calculation.
- Perfect cubes include numbers like 1 (since \(1 \times 1 \times 1\)), 8 (\(2 \times 2 \times 2\)), 27 (\(3 \times 3 \times 3\)), and so forth.
- To factor a larger cube, identify smaller perfect cube components within it that can be broken down into simpler terms.
Other exercises in this chapter
Problem 73
Rationalize each denominator. All variables represent positive real numbers. $$ \frac{\sqrt[3]{4 a^{6}}}{\sqrt[3]{2 a^{5} b}} $$
View solution Problem 73
Simplify each expression, if possible. All variables represent positive real numbers. $$ \sqrt[6]{m^{11}} $$
View solution Problem 74
Explain how the distance formula and the Pythagorean theorem can be used to show that a triangle with vertices \((2,3),(-3,4)\) and \((1,-2)\) is a right triang
View solution Problem 74
Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{b+7}-\sqrt{b-5}=2 $$
View solution