Question

Please help me/:
How can you write this expression with a rationalized denominator?
(2+cuberoot3)/cuberoot6

Answer 1

@rachelfreak123

Answer 2

so it looks like this right
\[\frac{ 2+\sqrt[3]{3} }{ \sqrt[3]{6} }\]

Answer 3

Right. Sorry, I wasn't quite sure how to type it.

Answer 4

no prob
so the trick with this is that you have cube roots, so you need to know how to simplify cube roots
i had a teacher teach me to put groups of three together to bring one out instead of trying to remember what all of the perfect cubes are.

So for example if you have
\[\sqrt[3]{54}\]
you break it into
\[\sqrt[3]{2\times3\times3\times3}\]
and since there are three 3's that would make a cube and so one can be pulled out to simplify
\[3\sqrt[3]{2}\]
does that make sense, trust me it is important to undertsand how to rationalize.

Answer 5

ok. Yes I understand that

Answer 6

so what you have to do is multiply the top and bottom of the fraction by another cube root so that you do not have a cube root in the bottom anymore.

Answer 7

Don't you multiply the top and bottom by the cube root of 6?

Answer 8

\[\frac{ 2+\sqrt[3]{3} }{ \sqrt[3]{6} } \times \frac{ \sqrt[3]{6 \times 6} }{ \sqrt[3]{6\times6} }\]

Answer 9

you have to have three 6's under the cube root to bring one out, so you need to multiply the top and bootom by cube root of 6 times 6

Answer 10

Ok then how do I do that? I'm sorry, I'm really stuck

Answer 11

the top is going to be tricky, so let's to the bottom first.

Answer 12

on the bottom you are multiplying so you can put them all under one cube root
\[\sqrt[3]{6\times6\times6}\]
so what will that be?

Answer 13

cube root of 216.....?

Answer 14

well if you have 3 of the same number under a cube root then you just have that number

Answer 15

so you do have cube root of 216, but if you took the cube root it would just be 6, so you do not have to mulitply it out, see....

Answer 16

Ohhh ok yes. I see

Answer 17

so on the bottome we will have 6, make sense?

Answer 18

yes

Answer 19

now on the top we are going to have to use the distrubutive property, blah

Answer 20

Ugh....

Answer 21

\[(2+\sqrt[3]{3}) \times \sqrt[3]{6\times6}\]
\[2\times \sqrt[3]{6\times6}\]
this will be the first term, there will not be much that we can do with it, except to multiply the 6s under the cube root
\[\sqrt[3]{3}\times \sqrt[3]{6\times6}\]
the second term is the tricky on, we are going to have to break the 6's down into 2 times 3

Answer 22

since they are all cube roots we can write them all under one
\[\sqrt[3]{3\times6\times6}\]
but we want to break the 6s down
\[\sqrt[3]{3\times2\times3\times2\times3}\]

Answer 23

still with me?

Answer 24

yes. .

Answer 25

ok, so now that we have it all broken down uner the cube root we look to see if we have any number repeated three times, do you see a number three times under that cube root?

Answer 26

So the answer will be (2cuberoots36+3cuberoots4)/6???

Answer 27

yes!

Answer 28

YAY!! Thank you SOOO much!!!!