## anonymous 5 years ago rationalizing denominators, the cube root of 2y squared divided by the cube root of 9x squared

1. anonymous

$\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}$ $\implies\frac{\sqrt[3]{18yx^2}}{9x^2}$

2. amistre64

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

3. amistre64

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

4. anonymous

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

5. amistre64

no prob; happens to me all the time :)

6. amistre64

trig test in 20mins...yawn :)

7. anonymous

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

8. amistre64

denom will rationalize to 3x

9. amistre64

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

10. anonymous

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

11. anonymous

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

12. amistre64

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

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

14. amistre64

15. anonymous

how?

16. amistre64

Ciao :)

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

18. anonymous

@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).

My first post was just seeking clarification.

20. anonymous

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

I am talking about the original problem posted by abdon

What is squared the 2y or just the y

23. anonymous

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

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

25. anonymous

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

26. anonymous

$got class ,thanks still confusing$

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