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zepdrix
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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? :)

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

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

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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0okay i see what ur sayying

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

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

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

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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0okay thankyou(:could u help me with another?

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

zepdrix
 2 years 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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