Question

rationalizing denominators, the cube root of 2y squared divided by the cube root of 9x squared

Answer 1

\[\frac{\sqrt[3]{2y}}{\sqrt[3]{9x^2}} \times \frac{\sqrt[3]{9x^2}}{\sqrt[3]{9x^2}}\]\[\implies\frac{\sqrt[3]{2y * 9x^2}}{9x^2}\] \[\implies\frac{\sqrt[3]{18yx^2}}{9x^2}\]

Answer 2

cbrt(9x^2)^2 does not equal 9x^2...

Answer 3

cbrt((3^2 x^2) times (cbrt(3x)) = 3x

Answer 4

Sorry, I was thinking of a regular square root, thanks for catching that amistre. :P

Answer 5

no prob; happens to me all the time :)

Answer 6

trig test in 20mins...yawn :)

Answer 7

So the denominator should be \[\sqrt[3]{81x^4}\].

Answer 8

denom will rationalize to 3x

Answer 9

cbrt(3^2 x^2 3 x) = cbrt(3^3 x^3) = 3x

Answer 10

\[thats what I got but the answer \in the back of the book says cube \root of6xy squred divided by 3x\]

Answer 11

You're right. :P So the numerator should, in fact, be \[\sqrt[3]{6yx}\] and the denominator\[3x.\]

Answer 12

top:

cbrt(2y) cbrt(3x) = cbrt(6xy) yep; checks out on my "mr proffesor"

Answer 13

Is it\[\sqrt[3]{(2y)2}\]
or\[\sqrt[3]2y{2}\]

Answer 14

how?

Answer 15

Ciao :)

Answer 16

The first expression is sort of confusing as it is written.

Answer 17

@abdon: If you multiply the denominator by cbrt(3x) you get cbrt(27*x^3) which simplifies to 3x. Now, because you multiplied the denominator by that factor, you have to multiply the numerator by the same factor, and you get cbrt(6xy).

Answer 18

My first post was just seeking clarification.

Answer 19

Ohhhhhhh, sorry, it is y^2...but that stays constant throughout, didn't see that before.

Answer 20

I am talking about the original problem posted by abdon

Answer 21

What is squared the 2y or just the y

Answer 22

I believe it was \[\sqrt[3]{2y^2}/\sqrt[3]{9x^2}.\]

Answer 23

OK. Thanks, now I will see if I get the same as you.

Answer 24

Sorry for having such broken up responses, here's the condensed form of what I got:\[\frac{\sqrt[3]{6xy^2}}{3x}.\]

Answer 25

\[got class ,thanks still confusing\]