How can you write the expression with rationalized denominator?

- anonymous

How can you write the expression with rationalized denominator?

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- schrodinger

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- anonymous

@nooce

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- anonymous

@zepdrix

- zepdrix

Oh sorry I'm helping someone else, forgot about this one c: hehe
glad you pinged me XD

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## More answers

- zepdrix

So we have an irrational number in the denominator. To fix that, let's ummm

- zepdrix

\[\large \frac{1}{\sqrt2}\]Here's a quick example, when dealing with a SQUARE (2nd) root, we can multiply the top and bottom by \(\huge \frac{\sqrt2}{\sqrt2}\) and it will end up changing the denominator to a rational number, namely, 2.
But in this problem we have a CUBE (3rd) root, so we have to do something a lil bit fancier.

- zepdrix

To get a 6 in the denominator we'll need to multiply the top and bottom by \(\huge \frac{6^{2/3}}{6^{2/3}}\)

- anonymous

okay....

- zepdrix

\[\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}\]Understand the fractional exponent notation ok? :)

- anonymous

ohhh okay (: i get ya

- zepdrix

\[\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)\]Understand how the bottom will simplify? c:

- anonymous

i believe so lol

- anonymous

now what?

- zepdrix

When we MULTIPLY terms with similar bases, we ADD the exponents.
So in the bottom we'll get,\[\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}\]

- zepdrix

\[\huge = \qquad 6^{3/3} \qquad = 6\]

- anonymous

so the bottom is 6 lol

- zepdrix

Yah :3

- anonymous

so now the top?

- zepdrix

Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)

- anonymous

i could give u the possible answers..... they all have 6 under them

- zepdrix

oh ok that might help ^^

- anonymous

okay hold on

- zepdrix

Oh ok I see what they want us to do.
So now that we've taken care of the bottom, we'll write the \(\large 6^{2/3}\) like this \(\large \sqrt[3]{6^2}\) before we distribute it to the top.

- zepdrix

\[\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}\]And I guess they want us to simplify a bit from here.

- anonymous

okay i see what ur sayying

- zepdrix

\[\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}\]

- zepdrix

This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.

- zepdrix

The following numbers are perfect cubes,
8, 27, 64, 125.
Because if you take the cube root of any of those numbers, they'll give you a nice clean value.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]

- zepdrix

It turns out that 108 is divisible by 27!\[\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}\]

- anonymous

soooo

- zepdrix

Ah sorry lost my connection D:

- anonymous

27 isnt a possible answer....

- zepdrix

27 is a perfect cube! So let's take the cube root of 27.

- zepdrix

\[\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3\]

- zepdrix

This is a really awful problem. Is this for algebra or something? D:

- anonymous

algabra 2 :/

- zepdrix

So it looks like it isssss probablyyyyyyy .... d.
I'm like .. 65% sure :D lol

- zepdrix

It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.

- anonymous

okay thankyou(:could u help me with another?

- anonymous

it's not like that one i promise lol

- zepdrix

Close this thread.
Start a new one with your new question.
Type @zepdrix somewhere in the comments and I'll try to take a look if I have time ^^

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