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Chapter 0: Problem 85

\(83-88=\) Rationalize the denominator. $$ \begin{array}{llll}{\text { (a) } \frac{1}{\sqrt[3]{4}}} & {\text { (b) } \frac{1}{\sqrt[4]{3}}} & {\text { (c) } \frac{8}{\sqrt[5]{2}}}\end{array} $$

Short Answer

Expert verified
(a) \(\frac{\sqrt[3]{16}}{4}\), (b) \(\frac{\sqrt[4]{27}}{3}\), (c) \(4\cdot \sqrt[5]{16}\).

Step by step solution

01

Rationalizing the denominator for (a)

We have \(\frac{1}{\sqrt[3]{4}}\). To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt[3]{16}\), because \(4 \times 16 = 64\), which is a perfect cube. This gives \(\frac{\sqrt[3]{16}}{\sqrt[3]{64}} = \frac{\sqrt[3]{16}}{4}\).
02

Rationalizing the denominator for (b)

We have \(\frac{1}{\sqrt[4]{3}}\). Multiply the numerator and the denominator by \(\sqrt[4]{27}\), because \(3 \times 27 = 81\), which is a perfect fourth power. This results in \(\frac{\sqrt[4]{27}}{\sqrt[4]{81}} = \frac{\sqrt[4]{27}}{3}\).
03

Rationalizing the denominator for (c)

We have \(\frac{8}{\sqrt[5]{2}}\). To rationalize, multiply the numerator and the denominator by \(\sqrt[5]{16}\), since \(2 \times 16 = 32\), a perfect fifth power. This leads us to \(\frac{8\cdot \sqrt[5]{16}}{2} = 4\cdot \sqrt[5]{16}\).

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Key Concepts

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

Perfect Cube
Understanding the concept of a perfect cube is important when dealing with cube roots in mathematics. A perfect cube is an integer that is the cube of another integer. In simpler terms, it's a number that can be expressed as the product of three identical factors of a number. For example, 27 is a perfect cube because it can be expressed as
  • \[ 3 imes 3 imes 3 = 27 \]
  • which is \(3^3\).
When simplifying expressions like \(\frac{1}{\sqrt[3]{4}}\), we aim to remove the cube root in the denominator. To do this, find a number that multiplied by 4 becomes a perfect cube. In this case, it's 16, because
  • \[ 4 \times 16 = 64 \]
  • and 64 is \(4^3\).
Thus, multiply both the numerator and the denominator by \(\sqrt[3]{16}\) to rationalize the cube root.
Perfect Fourth Power
Next, let鈥檚 dive into the perfect fourth power concept, especially as seen in fourth roots. A perfect fourth power is an integer that results from raising a number to the power of four. It is like squaring a square. For example, 16 is a perfect fourth power because
  • \[ 2^4 = 16 \]
  • or \(2 \times 2 \times 2 \times 2\).
In exercises like \(\frac{1}{\sqrt[4]{3}}\), we aim to find a number that makes 3 into a perfect fourth power when multiplied. This number is 27, as
  • \[ 3 \times 27 = 81 \]
  • where 81 is \(3^4\).
So, multiplying both numerator and denominator by \(\sqrt[4]{27}\) helps eliminate the fourth root from the denominator.
Perfect Fifth Power
Perfect fifth power is a similar concept, but it relates to raising a number to the power of five. It鈥檚 more extensive than the previous cases but follows a similar logic. A number like 32 is a perfect fifth power because
  • \[ 2^5 = 32 \]
  • or \(2 \times 2 \times 2 \times 2 \times 2 \).
In simplifying an expression like \(\frac{8}{\sqrt[5]{2}}\), find a harmonizing number such that 2 multiplied by it becomes a perfect fifth power. In this instance, it's 16, since
  • \[ 2 \times 16 = 32 \]
  • which is \(2^5\).
Therefore, multiplying both the top and bottom by \(\sqrt[5]{16}\) ensures the fifth root vanishes from the denominator.

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Most popular questions from this chapter

Perform the addition or subtraction and simplify. $$ \frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}} $$

State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{-a}{b}=-\frac{a}{b} $$

Perform the multiplication or division and simplify. $$ \frac{x^{2}-x-6}{x^{2}+2 x} \cdot \frac{x^{3}+x^{2}}{x^{2}-2 x-3} $$

Perform the multiplication or division and simplify. $$ \frac{t-3}{t^{2}+9} \cdot \frac{t+3}{t^{2}-9} $$

Perform the addition or subtraction and simplify. $$ \frac{1}{x+3}+\frac{1}{x^{2}-9} $$
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