Chapter 9.1, Problem 68E

a.

To determine

To approximate the given cube root numbers to nearest integers.

Answer to Problem 68E

The cube root numbers to nearest integers of given numbers as: $\sqrt[3]{3}=\text{ 2, }\sqrt[3]{55},=\text{ 4, }\sqrt[3]{22}=3.$

Explanation of Solution

Given: The given numbers are. $\sqrt[3]{3}\text{, }\sqrt[3]{55}\text{, }\sqrt[3]{22}.$

Calculation:

The cube root of a number $n$ is consider as $m$ . The table values for cube root of given numbers as:

  $\begin{array}{cc}n& m=\sqrt[3]{n}\\ 3& \sqrt[3]{3}={\left(3\right)}^{1/3}=1.44\\ 55& \sqrt[3]{55}={\left(55\right)}^{1/3}=3.8\\ 22& \sqrt[3]{22}={\left(22\right)}^{1/3}=2.8\end{array}$Substitute $n$ value in equation $m$ .

1. Let the given number: $\sqrt[3]{3}$ : The perfect cube closest to, but less than $3$ is $1$ . And The perfect cube closest to, but greater than $3$ is $8$ .

  $\begin{array}{l}So3\text{ is between 1 and 2}\text{.}\\ \text{The statement can be expressed by the compound inequality 1 < 3 <4}\text{.}\\ \text{1 < 3 < 2 Identify perfect cubes closest to 3}\text{.}\\ \sqrt[3]{1}<\sqrt[3]{3}<\sqrt[3]{8}\text{ Take positive cube root of each number}\text{.}\\ \text{1 < }\sqrt[3]{3}\text{ < 2 Evaluate cube root of each perfect cube}\text{.}\\ The\text{ average of 1 and 2 is 1}\text{.5 and (1}{\text{.5)}}^3=\text{ 3}\text{.375}.\text{}Because\text{ 3 > 2}\text{. }\\ \text{That means }\sqrt[3]{3}\text{ is closer to 2 than to1}\text{. So, to the nearest integer, }\sqrt[3]{3}=2.\end{array}$

2. Let the given number: $\sqrt[3]{55}$ : The perfect cube closest to, but less than $55$ is $3$ . And The perfect cube closest to, but greater than $3$ is $4$ .

  $\begin{array}{l}So\text{ 55 is between 3 and 4}\text{.}\\ \text{The statement can be expressed by the compound inequality 3 < 55 < 4}\text{.}\\ \text{3 < 55 < 4 Identify perfect cubess closest to 55}\text{.}\\ \sqrt[3]{27}<\sqrt[3]{55}\text{<}\sqrt[3]{64}\text{ Take positive cube root of each number}\text{.}\\ \text{3 < }\sqrt[3]{55}\text{ < 4 Evaluate cube root of each perfect cube}\text{.}\\ The\text{ average of 3 and 4 is 3}\text{.5 and (3}{\text{.5)}}^3=\mathrm{42.875.}\text{}Because\text{ 55 > 42}\text{.875}\text{. }\\ \text{That means }\sqrt[3]{55}\text{ is closer to 4 than to 3}\text{. So, to the nearest integer, }\sqrt[3]{55}=4\end{array}$

3. Let the given number: $\sqrt[3]{22}$ : The perfect cube closest to, but less than $22$ is $8$ . And The perfect cube closest to, but greater than $22$ is $27$ .

  $\begin{array}{l}So\text{ 22 is between 2 and 3}\text{.}\\ \text{The statement can be expressed by the compound inequality 2 < 22 < 3}\text{.}\\ \text{2 < 22 < 3 Identify perfect cubes closest to 22}\text{.}\\ \sqrt[3]{8}<\sqrt[3]{22}<\sqrt[3]{27}\text{ Take positive cube root of each number}\text{.}\\ \text{2 < }\sqrt[3]{22}\text{ < 3 Evaluate cube root of each perfect cube}\text{.}\\ The\text{ average of 2 and 3 is 2}\text{.5 and (2}{\text{.5)}}^3=\mathrm{15.625.}\text{}Because\text{ 22 > 15}\text{.625}\text{. }\\ \text{That means }\sqrt[3]{22}\text{ is closer to 3 than to 2}\text{. So, to the nearest integer, }\sqrt[3]{22}=\text{ 3}\text{.}\end{array}$

b.

To determine

Explain about cube root of negative numbers.

Answer to Problem 68E

Yes. It is possible to find the cube root of negative numbers. The odd number of negative numbers gives the result is negative.

Explanation of Solution

Given: Take $\sqrt[3]{\left(-8\right)}$

Calculation:

Yes. It is possible to find the cube root of a negative number.

Consider, the cube root of $-8=\sqrt[3]{\left(-8\right)}={[(-2^3)]}^{1/3}=-2$ .

Because the multiplication of $\left(-2\right)\left(-2\right)\left(-2\right)=-8$ . That means when multiply an odd number of negative numbers, the result is negative.

c.

To determine

To solve the given equation $x^3=125$ .

Answer to Problem 68E

The solution for the given equation is $x=5$ .

Explanation of Solution

Given: $x^3=125$

Calculation: $\text{The given equation can be written as: }x=\sqrt[3]{125}=5$

Let the given number: $\sqrt[3]{125}$ : The perfect cube closest to $125$ is $5$ .

  $\begin{array}{l}\sqrt[3]{125}\text{ = 5 Identify perfect cubes closest to 22}\text{.}\\ \sqrt[3]{125}\text{=5 Take positive cube root of each number}\text{.}\\ The{\text{ value is 5 and (5)}}^3=125.\text{}\\ \text{That means }\sqrt[3]{125}\text{ is equal to 5}\text{. So the equation could be evaluated as x= }\sqrt[3]{125}=\text{ 5}\text{.}\end{array}$