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Q18.

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Found in: Page 266

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Radicals and Exponents Evaluate each expression.a. ${\mathbf{\left(}\mathbf{-}\mathbf{5}\mathbf{\right)}}^{\mathbf{3}}$b. role="math" localid="1648704247043" $\mathbf{-}{\mathbf{5}}^{\mathbf{3}}$c.

a) The value of ${\left(-5\right)}^{3}$ is $\mathbf{-}\mathbf{125}$.

b) The value of $-{5}^{3}$ is $\mathbf{-}\mathbf{125}$.

c) The value of is 4.

See the step by step solution

## Part a Step 1. Given information.

The given expression is ${\left(-5\right)}^{3}$.

## Part a  Step 2. Write the concept.

The exponential expression “$a$ to the power $n$” is defined as:

${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$

## Part a  Step 3. Determine the value of the expression.

The given expression can be written as:

$\begin{array}{l}{\left(-5\right)}^{3}=\left(-5\right)×\left(-5\right)×\left(-5\right)\\ =-125\end{array}$

Thus, the value of ${\left(-5\right)}^{3}$ is $\mathbf{-}\mathbf{125}$.

## Part b Step 1. Given information.

The given expression is $-{5}^{3}$.

## Part b  Step 2. Write the concept.

The exponential expression “$a$ to the power $n$” is defined as:

${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$

## Part b  Step 3. Determine the value of the expression.

The given expression can be written as:

$\begin{array}{l}-{5}^{3}=-\left(5×5×5\right)\\ =-125\end{array}$

Thus, the value of $-{5}^{3}$ is $\mathbf{-}\mathbf{125}$.

## Part c Step 1. Given information.

The given expression is .

## Part c  Step 2. Write the concept.

The exponential expression “$a$ to the power $n$” is defined as:

${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$

## Part c  Step 3. Determine the value of the expression.

The given expression can be written as:

$\begin{array}{l}{\left(-5\right)}^{2}\cdot {\left(\frac{2}{5}\right)}^{2}=\left(-5\right)×\left(-5\right)×\left(\frac{2}{5}\right)×\left(\frac{2}{5}\right)\\ =\left(25\right)×\left(\frac{4}{25}\right)\\ =4\end{array}$

Thus, the value of is 4.