Problem 5 - Elementary and Intermediate Algebra

Question

Fill in the blanks. To obtain a ____________ -cube radicand in the denominator of \(\frac{\sqrt[3]{7}}{\sqrt[3]{5 n}}\) we multiply the fraction by \(\frac{\sqrt[3]{25 n^{2}}}{\sqrt[3]{25 n^{2}}}\).

Step-by-Step Solution

Answer

perfect

Step 1: Identifying the Original Fraction

The original fraction given is \(\frac{\sqrt[3]{7}}{\sqrt[3]{5n}}\). Our goal is to modify the denominator so that it becomes a perfect cube radicand.

Step 2: Understanding Perfect Cubes

A perfect cube is a number that can be expressed as \(k^3\), where \(k\) is an integer or expression. To make the denominator a perfect cube, both constants and variables must have exponents that are multiples of 3.

Step 3: Analyzing the Denominator

The denominator \(\sqrt[3]{5n}\) is not a perfect cube. To transform it, the radicand 5 needs to become \(5^3\), and \(n\) needs to become \(n^3\). This requires multiplying by \(5^2\) and \(n^2\), respectively.

Step 4: Deciding What to Multiply By

Since \(\sqrt[3]{5n} = \sqrt[3]{5^1 n^1}\), we need \(5^2\) and \(n^2\) to make \(\sqrt[3]{5^3n^3}\). Therefore, the fraction to multiply by is \(\frac{\sqrt[3]{25n^2}}{\sqrt[3]{25n^2}}\).

Step 5: Making the Denominator a Perfect Cube

Multiplying \(\sqrt[3]{5n}\) by \(\sqrt[3]{25n^2}\) gives \(\sqrt[3]{5^3n^3}\), which is \((5n)\) to the power of 3. Thus, \(\sqrt[3]{25n^2}\) ensures the denominator becomes a perfect cube.

Key Concepts

Perfect Cubes

In mathematics, a perfect cube is a number that results from raising an integer or expression to the power of three. For example, 8 is a perfect cube because it can be written as \(2^3 = 8\).Numbers such as 27, 64, and 125 are perfect cubes because you can express them as \(3^3, 4^3, \text{and} 5^3\) respectively.In the context of algebraic expressions, any variable expressed as, say, \(x^3\) is also a perfect cube.

To determine if a number or an expression is a perfect cube or to transform it into one, ensure that all parts of the number (including its prime factors) and variables have exponents that are multiples of three.This is useful when you need to simplify radical expressions, especially those involving cube roots.For instance, in the provided exercise, the denominator originally was \(\sqrt[3]{5n}\), and we needed it to become a perfect cube.Thus, we adjusted the factors inside the radical so that their exponents are all multiples of 3.

Rationalizing Denominators

Rationalizing the denominator is a technique used to eliminate radicals from the denominator of a fraction.This makes the fraction easier to work with, particularly in the context of simplification or further operations.

When dealing with cube roots, you can rationalize by multiplying both the numerator and the denominator by a suitable factor that turns the latter into a perfect cube.

• Identify what is missing in the denominator to turn it into a perfect cube, like finding the missing factors for the radical's components.
• Multiply both the numerator and the denominator by what's needed.
• This common strategy leaves the fraction mathematically equivalent to the original while having a 'tidy' denominator.

In the exercise, the denominator started as \(\sqrt[3]{5n}\). By multiplying it with \(\sqrt[3]{25n^2}\), we transformed the denominator into \(\sqrt[3]{5^3n^3} = (5n)\) cubed, thereby rationalizing it.

Cube Roots

A cube root is the inverse operation of cubing a number or an expression. When you cube a number, you multiply it by itself three times.So, taking a cube root gives you the original number before it was cubed.

Symbolically, the cube root of a number \(x\) is represented as \(\sqrt[3]{x}\).It asks the question, "What number multiplied by itself three times equals \(x\)?"For instance, \(\sqrt[3]{27} = 3\) because \(3^3 = 27\).

• Cube roots are particularly interesting when simplifying radical expressions.
• Understanding cube roots allows us to manipulate and simplify expressions further, especially when working with equations or fractions.
• Solving \(\sqrt[3]{5n}\) implies finding numbers \(a\) and \(b\) such that \(a^3 = 5n\), which often requires adjusting the factors inside the radical.

By using these principles, it's easier to handle and solve problems that involve cube roots, just like in the exercise where rationalizing the denominator involved employing cube roots effectively.