Question

Simplify:

$\left(a\right)\sqrt{\frac{54u^7}{v^8}}\left(b\right)\sqrt[3]{\frac{40r^3}{s^6}}\left(c\right)\sqrt[4]{\frac{162m^{14}}{n^{12}}}.$

Short answer

Part a. After simplifying we get,$\frac{3u^3\sqrt{6u}}{v^4}.$

Part b. After simplifying we get,$\frac{2r\sqrt[3]{5}}{s^2}.$

Part c. After simplifying we get,$\frac{3m^3\sqrt[4]{2m^2}}{n^3}.$

Step-by-step solution

Part a Step 1. Given Information 

The given expression is$\sqrt{\frac{54u^7}{v^8}}.$

We have to simplify it.

Part a Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

Thus, we write $\sqrt{\frac{54u^7}{v^8}}$ as $\frac{\sqrt{54u^7}}{\sqrt{v^8}}.$

Part a Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get,

$$\begin{gathered} \frac{\sqrt{54u^7}}{\sqrt{v^8}} \\ =\frac{\sqrt{{\left(3u^3\right)}^2}·\sqrt{6u}}{\sqrt{{\left(v^4\right)}^2}} \\ =\frac{3u^3\sqrt{6u}}{v^4} \end{gathered}$$

Part b Step 1. Given Information 

The given expression is$\sqrt[3]{\frac{40r^3}{s^6}}.$

We have to simplify it.

Part b Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

Thus, we write $\sqrt[3]{\frac{40r^3}{s^6}}$as $\frac{\sqrt[3]{40r^3}}{\sqrt[3]{s^6}}.$

Part b Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get,

$$\begin{gathered} \frac{\sqrt[3]{40r^3}}{\sqrt[3]{s^6}} \\ =\frac{\sqrt[3]{{\left(2r\right)}^3}·\sqrt[3]{5}}{\sqrt[3]{{\left(s^2\right)}^3}} \\ =\frac{2r\sqrt[3]{5}}{s^2} \end{gathered}$$

Part c Step 1. Given Information 

The given expression is$\sqrt[4]{\frac{162m^{14}}{n^{12}}}.$

We have to simplify it.

Part c Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

Thus, we write$\sqrt[4]{\frac{162m^{14}}{n^{12}}}$as$\frac{\sqrt[4]{162m^{14}}}{\sqrt[4]{n^{12}}}.$

Part c Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get,

$$\begin{gathered} \frac{\sqrt[4]{162m^{14}}}{\sqrt[4]{n^{12}}} \\ =\frac{\sqrt[4]{{\left(3m^3\right)}^4}·\sqrt[4]{2m^2}}{\sqrt[4]{{\left(n^3\right)}^4}} \\ =\frac{3m^3\sqrt[4]{2m^2}}{n^3} \end{gathered}$$