Multiply:  $2\sqrt{3}\cdot4\sqrt{27}$  

Here's a "proof," using nothing but radicals, that  $-1\ =\ 1$ :

 $-1\ =\sqrt{-1}\cdot\sqrt{-1}\ =\ \sqrt{-1\cdot-1}$   
 $=\sqrt{1\ }=\ 1$  

Which step contains the mistake? Explain why. Consider the first equals sign to be the first step, the second equals sign to be the second step, and so on.

Add:  $3\sqrt{7}\ -2\sqrt{28}+10\sqrt{4}$ 

Add: $6\sqrt{24}-24\sqrt{6}$  

Which number is larger: $30\sqrt{20}\$  or  $15\sqrt{80}$ ?