<s>[INST] Applying difference of cubes to cube roots 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. 
$$\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}.$$
 [/INST] 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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