Question

how can you write the expression with a rationalized denominator? ^3sqrt2/^3sqrt4 ??

Answer 1

\(\large\color{#000000 }{ \displaystyle \frac{\sqrt[3]{2} }{\sqrt[3]{4}} }\) like this?

Answer 2

Yes

Answer 3

Example problem 1:

\(\large\color{#000000 }{ \displaystyle \frac{\sqrt[3]{3} }{\sqrt[3]{9}} }\)

I know that,
\(\large\color{#000000 }{ \displaystyle \sqrt[3]{9} \times \sqrt[3]{3}=\sqrt[3]{27}=3 }\)

So, let's multiply times \(\large\color{#000000 }{ \displaystyle \sqrt[3]{3}}\) on top and bottom.

\(\large\color{#000000 }{ \displaystyle \frac{\sqrt[3]{3} \times \sqrt[3]{3} }{\sqrt[3]{9}\times \sqrt[3]{3}} =\frac{\sqrt[3]{9}}{3} }\)

Answer 4

Ok

Answer 5

Example problem 2:

\(\color{#000000 }{ \displaystyle \frac{\sqrt[3]{4} }{\sqrt[3]{16}} }\)

I know that,
\(\color{#000000 }{ \displaystyle \sqrt[3]{16} \times \sqrt[3]{4}= \sqrt[3]{4^3}=4 }\)

So, let's multiply times \(\large\color{#000000 }{ \displaystyle \sqrt[3]{4}}\) on top and bottom.

\(\color{#000000 }{ \displaystyle \frac{\sqrt[3]{4} \times \sqrt[3]{4} }{\sqrt[3]{16}\times \sqrt[3]{4}} =\frac{\sqrt[3]{16}}{4} }\)

Answer 6

you are just multiplying (on top and bottom) times something that would make the denominator into integer, basically/

Answer 7

Ooh ok. So you would multiply the top by the bottom and get ^3sqrt 8

Answer 8

Wow

Answer 9

Wouldn't it just be 8?

Answer 10

Well, you are multiplying by \(\sqrt[3]{2}\), but not because it is the top, RATHER you are doing because that would (indeed) get you \(\sqrt[3]{8}=2\).

And, you are multiplying times \(\sqrt[3]{2}\) on TOP and BOTTOM.

Answer 11

Oh ok.