Q. 76 - Intermediate Algebra

Question

In the following exercises, simplify using absolute value signs as needed. $\sqrt{96r^3{s}^3},\sqrt[3]{80x^7{y}^6},\sqrt[4]{80x^8{y}^9}$

Step-by-Step Solution

Answer

The expression on simplification is $4\left|rs\right|\sqrt{6rs},2x^2{y}^2\sqrt[3]{10x},2x^2{y}^2\sqrt[4]{5y}$ respectively

Step 1: Given information

Given the expression $\sqrt{96r^3{s}^3},\sqrt[3]{80x^7{y}^6},\sqrt[4]{80x^8{y}^9}$

Step 2: Using the product property

Using the product property, we get

$$\begin{gathered} \sqrt{96r^3{s}^3}=\sqrt{16r^2{s}^2·6rs} \\ \sqrt{96r^3{s}^3}=\sqrt{(4rs{)}^2·6rs} \\ \sqrt{96r^3{s}^3}=\sqrt{(4rs{)}^2}·\sqrt{6rs} \\ \sqrt{96r^3{s}^3}=4\left|rs\right|\sqrt{6rs} \end{gathered}$$

Step 3: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[3]{80x^7{y}^6}=\sqrt[3]{8x^6{y}^6·10x} \\ \sqrt[3]{80x^7{y}^6}=\sqrt[3]{{\left(2x^2{y}^2\right)}^3·10x} \\ \sqrt[3]{80x^7{y}^6}=\sqrt[3]{{\left(2x^2{y}^2\right)}^3}·\sqrt[3]{10x} \\ \sqrt[3]{80x^7{y}^6}=2x^2{y}^2\sqrt[3]{10x} \end{gathered}$$

Step 4: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[4]{80x^8{y}^9}=\sqrt[4]{16x^8{y}^8·5y} \\ \sqrt[4]{80x^8{y}^9}=\sqrt[4]{{\left(2x^2{y}^2\right)}^4·5y} \\ \sqrt[4]{80x^8{y}^9}=\sqrt[4]{{\left(2x^2{y}^2\right)}^4}·\sqrt[4]{5y} \\ \sqrt[4]{80x^8{y}^9}=2\left|x^2{y}^2\right|\sqrt[4]{5y} \\ \sqrt[4]{80x^8{y}^9}=2x^2{y}^2\sqrt[4]{5y} \end{gathered}$$