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

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

zepdrix
 one year ago
Best 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.

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

zepdrix
 one year ago
Best 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? :)

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

zepdrix
 one year ago
Best 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:

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best 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.

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

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

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

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

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

kandsbbf
 one year ago
Best ResponseYou've already chosen the best response.027 isnt a possible answer....

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.027 is a perfect cube! So let's take the cube root of 27.

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

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Close 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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