[\MATH]\frac{1}{((x)^(1/3)-(y)^(1/3))}[/MATH]
$\displaystyle \frac1{x^{1/3}-y^{1/3}}$
$\displaystyle =\frac1{x^{1/3}-y^{1/3}}\cdot\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}$
$\displaystyle =\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x-y}$
Note that the expression which I multiplied the numerator and denominator by is just the second half of the difference of cubes formula.
I've noticed... but I don't understand why you have to do that but rules are rules I guess... Thank you so much for your response!!! But how is that x-y in the denominator?