Question

Evaluate $\frac{a^2-b^2}{8a{b}^3}$for each value:

$ⓐa=1,b=-1ⓑa=\frac{1}{2},b=-1ⓒa=-2,b=1$

Short answer

$$\begin{gathered} ⓐ\mathrm{The}\mathrm{value}\mathrm{of}\frac{a^2-b^2}{8a{b}^3}\mathrm{for}a=1,b=-1\mathrm{is}0. \\ ⓑ\mathrm{The}\mathrm{value}\mathrm{of}\frac{a^2-b^2}{8a{b}^3}\mathrm{for}a=\frac{1}{2},b=-1\mathrm{is}\frac{3}{16}. \\ ⓒ\mathrm{The}\mathrm{value}\mathrm{of}\frac{a^2-b^2}{8a{b}^3}\mathrm{for}a=-2,b=1\mathrm{is}-\frac{3}{16}. \end{gathered}$$

Step-by-step solution

Step 1. Given Information 

Evaluate $\frac{a^2-b^2}{8a{b}^3}$for each value:

$ⓐa=1,b=-1ⓑa=\frac{1}{2},b=-1ⓒa=-2,b=1$

Part (a) Step 1. Finding the value of a2−b28ab3 for a=1, b=−1

Putting $a=1,b=-1$on the given expression

role="math" localid="1648381472744" $$\begin{gathered} \frac{a^2-b^2}{8a{b}^3}=\frac{{\left(1\right)}^2-{(-1)}^2}{8·1·{(-1)}^3} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{1-1}{8·1·(-1)} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{0}{-8} \\ \frac{a^2-b^2}{8a{b}^3}=0 \end{gathered}$$

Part (b) Step 1. Finding the value of a2−b28ab3 for a=12 , b=−1

Putting $a=\frac{1}{2},b=-1$on the given expression

$$\begin{gathered} \frac{a^2-b^2}{8a{b}^3}=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2-{(-1)}^2}{8·{\displaystyle \frac{1}{2}}·{(-1)}^3} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{{\displaystyle \frac{1}{4}}-1}{4·(-1)} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{{\displaystyle \frac{1}{4}-1·\frac{4}{4}}}{-4} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{{\displaystyle \frac{1}{4}-\frac{4}{4}}}{-4} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{{\displaystyle \frac{1-4}{4}}}{-4} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{-3}{4}·\frac{1}{-4} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{3}{16} \end{gathered}$$

Part (c) Step 1. Finding the value of a2−b28ab3 for a=−2, b=1

Putting $a=-2,b=1$on the given expression

role="math" localid="1648381736852" $$\begin{gathered} \frac{a^2-b^2}{8a{b}^3}=\frac{{(-2)}^2-{\left(1\right)}^2}{8·(-2)·{\left(1\right)}^3} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{4-1}{8·(-2)·1} \\ \frac{a^2-b^2}{8a{b}^3}=\frac{3}{-16} \\ \frac{a^2-b^2}{8a{b}^3}=-\frac{3}{16} \end{gathered}$$