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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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  • $\begingroup$ We have $x^3-y^3=(x-y)(x^2+xy+y^2)$. Let $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$. $\endgroup$ – André Nicolas Apr 17 '12 at 6:38
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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 $y=\sqrt{b}$.)

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