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Chapter 10: Problem 92

\(\sqrt[3]{\frac{64}{125}}\)

Short Answer

Expert verified
The cube root of \(\frac{64}{125}\) is \(\frac{4}{5}\).

Step by step solution

01

Identify the Problem

The problem is to find the cube root of the fraction \(\frac{64}{125}\).
02

Rewrite the Problem

Rewrite the problem as \(\sqrt[3]{\frac{64}{125}}\). This can be separated into two parts: \(\sqrt[3]{64}\) and \(\sqrt[3]{125}\).
03

Find the Cube Root of the Numerator

Determine \(\sqrt[3]{64}\). Since \(64 = 4^3\), \(\sqrt[3]{64} = 4\).
04

Find the Cube Root of the Denominator

Determine \(\sqrt[3]{125}\). Since \(125 = 5^3\), \(\sqrt[3]{125} = 5\).
05

Combine the Results

Combine the results from the numerator and the denominator: \(\frac{\sqrt[3]{64}}{\sqrt[3]{125}} = \frac{4}{5}\).
06

Simplify the Fraction

The simplest form of \(\frac{4}{5}\) is already given, so there is no further simplification needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cube Root
A cube root is a number that, when multiplied by itself twice, gives the original number. It's written as \(\root{3}{x}\) for a number x.
Taking the cube root of a fraction like \(\root{3}{\frac{64}{125}}\) involves finding the cube root of both the numerator and the denominator separately.
For example:
  • If you have \(\root{3}{64}\), you are looking for a number that when multiplied by itself twice equals 64. In this case, \(\root{3}{64} = 4\) because \({4 \times 4 \times 4 = 64}\).
  • Similarly, for \(\root{3}{125}\), since \({5 \times 5 \times 5 = 125}\), hence \(\root{3}{125} = 5\).
Thus, \(\root{3}{\frac{64}{125}}\) simplifies to \(\frac{\root{3}{64}}{\root{3}{125}} = \frac{4}{5}\).
Fraction Simplification
Simplifying fractions is about reducing them to their simplest form. This often involves finding the greatest common factor (GCF) of the numerator and denominator.
A fraction is in its simplest form when the numerator and denominator have no common factors other than 1. In our problem, \(\frac{4}{5}\) is already in its simplest form:
Both 4 and 5 are prime relative to each other (no common factors), so no further simplification is needed.
When finding cube roots of fractions, it helps to keep the end fraction in a simplified state. In this example:
  • Numerator: 4 is already simplified.
  • Denominator: 5 is also already simplified.
Thus, the result \(\frac{4}{5}\) is a simplified fraction.
Prime Factorization
Prime factorization involves breaking down a number into a product of its prime numbers. These are numbers that can only be divided by 1 and themselves.
This concept helps us understand whether a denominator or numerator can be simplified. Let's apply prime factorization to understand cube roots:
Example Numerator: Prime Factorization of 64
  • 64 = 2 脳 2 脳 2 脳 2 脳 2 脳 2 (or \(2^6\))
  • We group these into three sets of 2: \(2^3 脳 2^3 = {(2^3)}^2 = 4^2\)
Since cube root reverses the cube effect, \root{3}{64} = 4\.
Example Denominator: Prime Factorization of 125
  • 125 = 5 脳 5 脳 5 (or \(5^3\))
  • It鈥檚 already grouped in sets of three: \root{3}{5^3} = 5\.
Prime factorization helps confirm the cube roots we've found and ensures accurate simplifications.

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Most popular questions from this chapter

List all of the following sets to which each number belongs. A number may belong to more than one set. real numbers pure imaginary numbers nonreal complex numbers complex numbers $$ 3+5 i $$

List all of the following sets to which each number belongs. A number may belong to more than one set. real numbers pure imaginary numbers nonreal complex numbers complex numbers $$ -7 i $$

Work each problem. Replace \(a\) with 3 and \(b\) with 4 to show that, in general, $$ \sqrt{a^{2}+b^{2}} \neq a+b $$

Simplify \(-\sqrt[4]{512}\)

Graph each circle. Identify the center and the radius. \(x^{2}+y^{2}=4\)
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