Q. 8.79 - Intermediate Algebra

Question

Simplify:

1. $\left(3\sqrt{2}\right)\left(2\sqrt{20}\right)$
   
2. $\left(2\sqrt[3]{18}\right)(-3\sqrt[3]{6})$
   

Step-by-Step Solution

Answer

Part a. $12\sqrt{10}$

Part b. $-18\sqrt[3]{4}$

Part (a) Step 1. Multiply the radicals and simplify

Using the product property of roots that is $\sqrt[n]{a}·\sqrt[n]{b}=\sqrt[n]{ab}$, the expression can be simplified to

$\begin{array}{rcl}\left(3\sqrt{2}\right)\left(2\sqrt{20}\right)& =& (3·2)(\sqrt{2}·\sqrt{20})\\ & =& 6\sqrt{2·20}\\ & =& 6\sqrt{40}\end{array}$

Now, the radical can be simplified as

$\begin{array}{rcl}6\sqrt{40}& =& 6\sqrt{4·10}\\ & =& 6\sqrt{4}·\sqrt{10}\\ & =& 6·2·\sqrt{10}\\ & =& 12\sqrt{10}\end{array}$

Part (b) Step 1. Multiply the radicals and simplify

Using the product property of roots that is $\sqrt[n]{a}·\sqrt[n]{b}=\sqrt[n]{ab}$, the expression can be simplified to

$\begin{array}{rcl}\left(2\sqrt[3]{18}\right)(-3\sqrt[3]{6})& =& 2·(-3)·(\sqrt[3]{18}·\sqrt[3]{6})\\ & =& -6\sqrt[3]{18·6}\\ & =& -6\sqrt[3]{108}\end{array}$

Now, the radical can be simplified as

$\begin{array}{rcl}-6\sqrt[3]{108}& =& -6\sqrt[3]{27·4}\\ & =& -6·\sqrt[3]{27}·\sqrt[3]{4}\\ & =& -6·3·\sqrt[3]{4}\\ & =& -18\sqrt[3]{4}\end{array}$