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When $\sqrt[3]{-128}$ is simplified, the result is $a\sqrt[3]{b}$, where $a$ is an integer, and $b$ is a positive integer. If $b$ is as small as possible, then what is $a+b$?

 Jun 30, 2015

Best Answer 

 #1
avatar+128474 
+5

3√[-128]  =

 

3√[ -1 * 64 * 2] =

 

3√ -1  *  3√64   *  3√2   =

 

-1 * 4  * 3√2   =

 

-4 3√2

 

So   a + b =   -4  + 2 =  -2

 

 Jun 30, 2015
 #1
avatar+128474 
+5
Best Answer

3√[-128]  =

 

3√[ -1 * 64 * 2] =

 

3√ -1  *  3√64   *  3√2   =

 

-1 * 4  * 3√2   =

 

-4 3√2

 

So   a + b =   -4  + 2 =  -2

 

CPhill Jun 30, 2015

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