+0  
 
0
42
4
avatar+2065 

I really don't understand these, please help >.<

RainbowPanda  Oct 24, 2018
 #1
avatar+2065 
0

Second one only

RainbowPanda  Oct 24, 2018
 #2
avatar+2742 
+2

\(\dfrac{\sqrt[3]{4r^3}}{\sqrt[3]{9r}} = \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \\ \\ \text{This is really as far as it can be simplified}\\ \text{but it's common to want the denominator to be free of radicals}\\ \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} =\dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \dfrac{\sqrt[3]{9^2r^2}}{\sqrt[3]{9^2r^2}} = \dfrac{r\sqrt[3]{324r^2}}{\sqrt[3]{9^3r^3}}=\dfrac{r\sqrt[3]{324r^2}}{9r}= \dfrac{\sqrt[3]{324r^2}}{9}\)

Rom  Oct 24, 2018
 #3
avatar+2065 
+1

What about the 2? Where does that go?

RainbowPanda  Oct 24, 2018
 #4
avatar+2742 
+1

eh b****y h**l I missed it.

 

It goes in the denominator so the denominator becomes 18 rather than 9.

Rom  Oct 25, 2018

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