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

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

okay....

ohhh okay (: i get ya

i believe so lol

now what?

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

so the bottom is 6 lol

Yah :3

so now the top?

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

oh ok that might help ^^

okay hold on

okay i see what ur sayying

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

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

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

soooo

Ah sorry lost my connection D:

27 isnt a possible answer....

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

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

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

algabra 2 :/

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

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

okay thankyou(:could u help me with another?

it's not like that one i promise lol