I do not know where to start this

[John292]

Hello,

If someone can get me started with this I would be greatful.


(8x)/[2xy[sup:3huxetmw]2[/sup:3huxetmw]][sup:3huxetmw]1/3[/sup:3huxetmw]

 

[mmm4444bot]

:idea:

You could start by telling us what you've been asked to do with this expression.

 

[John292]

I am very sorry! They asked to rationalize the denominator and simplify. It is Y and not V as it shows up.

 

[Deleted member 4993]

I am very sorry! They asked to rationalize the denominator and simplify. It is Y and not V as it shows up.

To you - what does rationalizing the denominator mean?

 

[John292]

The answer is

4[4x[sup:t4uygghy]2[/sup:t4uygghy]y][sup:t4uygghy]1/3[/sup:t4uygghy]/y

 

[mmm4444bot]

We know what the answer is.

What we do not know is why you're stuck.

Subhotosh formed a specific question to determine whether or not you know the meaning of the phrase "rationalize the denominator".

You did not answer his question. Would you like me to start guessing ?

If so, then my first guess is that (1) you have some idea of what it means to rationalize a denominator, and (2) you do not recognize that the given denominator is a radical.

Based on this guess, I offer the following information (hoping that it helps you to get started).

(2xy^2)^(1/3) is a cube root.

Does seeing the given expression rewritten in radical form help you ?

$\displaystyle \frac{8x}{\sqrt[3]{2xy^2}}$

If not, then I will simply say that we rationalize the denominator in the given ratio by multiplying top and bottom by (2xy^2)^(2/3), followed by simplifying the result.

I welcome specific questions.

 

[John292]

The way you wrote the problem is the way it is written in the text I have. I do not know how to write it the way you did with the editor. Please forgive me.

I think I understand what you say. I give it a try

 

[BigGlenntheHeavy]

$\displaystyle Given: \ \frac{8x}{(2xy^2)^{1/3}}, \ Rationalize \ the \ denominator \ and \ simplify.$

$\displaystyle \frac{8x*(2xy^2)^{2/3}}{(2xy^2)^{1/3}*(2xy^2)^{2/3}} \ = \ \frac{8x(2xy^2)^{2/3}}{2xy^2} \ = \ \frac{4(2xy^2)^{2/3}}{y^2}$

$\displaystyle = \ \frac{4(2^2)^{1/3}(x^2)^{1/3}(y)(y)^{1/3}}{y^2} \ = \ \frac{4(4x^2y)^{1/3}}{y}$

$\displaystyle Note: \ Neither \ x \ nor \ y \ = \ 0, \ why?$