Question

In the following exercises, simplify

$$\begin{gathered} \left(a\right)\sqrt{80c^7}-\sqrt{20c^7} \\ \left(b\right)2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}} \end{gathered}$$

Short answer

Simplifying,

$$\begin{gathered} \left(a\right)\sqrt{80c^7}-\sqrt{20c^7}=2\sqrt{5}{c}^3\sqrt{c} \\ \left(b\right)2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}}=14\sqrt[4]{2}{r}^2\sqrt[4]{r^2} \end{gathered}$$

Step-by-step solution

Step 1. Given information

We have given,

$$\begin{gathered} \left(a\right)\sqrt{80c^7}-\sqrt{20c^7} \\ \left(b\right)2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}} \end{gathered}$$

Step 2. Concept 

Here we are going to factor the number inside the radical such that at least on of them is perfect square.

Also, $\sqrt[n]{a^n}=a$

Here we are going to apply radical rule,

$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

Part (a) Step 1. Explanation

We have,

$\sqrt{80c^7}-\sqrt{20c^7}$

Simplifying it,

$\sqrt{80c^7}-\sqrt{20c^7}=\sqrt{16\times 5}·\sqrt{c{\left(c^3\right)}^2}-\sqrt{4\times 5}·\sqrt{c{\left(c^3\right)}^2}$

$\Rightarrow \sqrt{80c^7}-\sqrt{20c^7}=4\sqrt{5}{c}^3\sqrt{c}-2\sqrt{5}{c}^3\sqrt{c}$

$\Rightarrow \sqrt{80c^7}-\sqrt{20c^7}=2\sqrt{5}{c}^3\sqrt{c}(2-1)$

$\Rightarrow \sqrt{80c^7}-\sqrt{20c^7}=2\sqrt{5}{c}^3\sqrt{c}$

Part (a) Step 2. Conclusion 

Hence, value of$\sqrt{80c^7}-\sqrt{20c^7}$is$2\sqrt{5}{c}^3\sqrt{c}$.

Part (b) Step 1. Explanation

We have,

$2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}}$

Simplifying it ahead,

localid="1663690168534" $$\begin{gathered} 2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}}=2\sqrt[4]{162}\sqrt[4]{r^{10}}+4\sqrt[4]{32}\sqrt[4]{r^{10}} \\ =2·\hspace{0.17em}3\sqrt[4]{2}{r}^2\sqrt[4]{r^2}+4·\hspace{0.17em}2\sqrt[4]{2}{r}^2\sqrt[4]{r^2} \\ =6\sqrt[4]{2}{r}^2\sqrt[4]{r^2}+8\sqrt[4]{2}{r}^2\sqrt[4]{r^2} \\ =14\sqrt[4]{2}{r}^2\sqrt[4]{r^2} \end{gathered}$$

Part (b) Step 2. Conclusion

Hence, value of$2\sqrt[4]{162r^{10}}+4\sqrt[4]{32r^{10}}$is$14\sqrt[4]{2}{r}^2\sqrt[4]{r^2}$.