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Applying difference of cubes to cube roots

I am a little confused with this question, as I am trying to rationalize the denominator, I realize that I would want to use the sum of cubes formula.

$ (x^3+y^3)= (x+y)(x^2-xy+y^2) $

The question being,

$$ \frac{1}{\sqrt[3] x + \sqrt[3] y} $$

When I apply the above sum of cubes formula, knowing that I have $ (x + y) $ portion already there as $ \sqrt[3] x + \sqrt[3] y $, my denominator still does not rationalize.

What is the best method for rationalizing this denominator?

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marked as duplicate by Arturo Magidin, William, Thomas, Noah Snyder, Guess who it is. Oct 5 '12 at 13:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This should have been added to your previous question, not made into a new question here. –  Arturo Magidin Apr 17 '12 at 16:51
If your denominator does not rationalize, then you aren't doing it right. We have:$$\frac{1}{\sqrt[3]{x}+\sqrt[3]{y}} = \frac{(\sqrt[3]{x})^2 - \sqrt[3]{x}\sqrt[3]{y}+(\sqrt[3]{y})^2}{(\sqrt[3]{x}+\sqrt[3]{y})((\sqrt[3]{x})^‌​2 - \sqrt[3]{x}\sqrt[3]{y}+(\sqrt[3]{y})^2)} = \frac{\sqrt[3]{x^2} - \sqrt[3]{xy} +\sqrt[3]{y^2}}{(\sqrt[3]{x})^3 - (\sqrt[3]{x})^2\sqrt[3]{y}+\sqrt[3]{x}(\sqrt[3]{y})^2 + (\sqrt[3]{x})^2\sqrt[3]{y} - \sqrt[3]{x}(\sqrt[3]{y})^2+(\sqrt[3]{y})^3} = \frac{\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}}{x+y}.$$ –  Arturo Magidin Apr 17 '12 at 16:51

1 Answer 1

The problem is confusing because there are two distinct $x$ and $y$. Rewrite the first part as $u^3+v^3 = (u+v)(u^2-uv+v^2)$.

Then let $u=\sqrt[3] x$ and $v=\sqrt[3] y$, we see that $x + y = (\sqrt[3] x + \sqrt[3] y) (u^2-uv+v^2)$, or $$\frac{1}{\sqrt[3] x + \sqrt[3] y} = \frac{u^2-uv+v^2}{x+y} = \frac{x^{2/3} - (xy)^{1/3} + y^{2/3}}{x+y}$$

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