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I am stumped as to why this application of the difference of cubes is valid... I am rationalizing the denominator. I don't understand the reasoning of why the difference of cubes formula is applicable to cubed roots, removing the root one gets an exponent of $a^{1/3}$ - I know how to simplify this expression, but I am hoping someone can help me along with the logic.

$\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}.$

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    We have $x^3-y^3=(x-y)(x^2+xy+y^2)$. Let $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$.2012-04-17

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You use the the formula $(x-y)(x^2+xy+y^2) = x^3-y^3$ with $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$. You already have one of the factors on the left hand side, so you multiply by the other factor (and cancel it out). If you have $\frac{1}{x-y}$ then you can transform it into $\frac{1}{x-y} = \frac{x^2+xy+y^2}{(x-y)(x^2+xy+y^2)} = \frac{x^2+xy+y^2}{x^3-y^3}.$

That is, you want to multiply the numerator and denominator by $\left( \sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}\right).$

(Just like to rationalize $\frac{1}{\sqrt{a}-\sqrt{b}}$you use the formula $(x-y)(x+y)=x^2-y^2$ with $x=\sqrt{a}$ and $y=\sqrt{b}$.)