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zepdrixBest ResponseYou've already chosen the best response.0
Oh sorry I'm helping someone else, forgot about this one c: hehe glad you pinged me XD
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So we have an irrational number in the denominator. To fix that, let's ummm
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \frac{1}{\sqrt2}\]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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
To get a 6 in the denominator we'll need to multiply the top and bottom by \(\huge \frac{6^{2/3}}{6^{2/3}}\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}\]Understand the fractional exponent notation ok? :)
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
ohhh okay (: i get ya
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)\]Understand how the bottom will simplify? c:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
When we MULTIPLY terms with similar bases, we ADD the exponents. So in the bottom we'll get,\[\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\huge = \qquad 6^{3/3} \qquad = 6\]
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
so the bottom is 6 lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
i could give u the possible answers..... they all have 6 under them
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
oh ok that might help ^^
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}\]And I guess they want us to simplify a bit from here.
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
okay i see what ur sayying
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
It turns out that 108 is divisible by 27!\[\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Ah sorry lost my connection D:
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
27 isnt a possible answer....
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
27 is a perfect cube! So let's take the cube root of 27.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
This is a really awful problem. Is this for algebra or something? D:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So it looks like it isssss probablyyyyyyy .... d. I'm like .. 65% sure :D lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
okay thankyou(:could u help me with another?
 one year ago

kandsbbfBest ResponseYou've already chosen the best response.0
it's not like that one i promise lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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 ^^
 one year ago
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