[\MATH]\frac{1}{((x)^(1/3)-(y)^(1/3))}[/MATH]

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- Jun 3rd 2012, 08:17 PMpurplec16Rationalize the denominator
[\MATH]\frac{1}{((x)^(1/3)-(y)^(1/3))}[/MATH]

- Jun 3rd 2012, 08:18 PMpurplec16Re: Rationalize the denominator
\frac{1}{((x)^(1/3)-(y)^(1/3))}

- Jun 3rd 2012, 08:58 PMReckonerRe: Rationalize the denominator
$\displaystyle \frac1{x^{1/3}-y^{1/3}}$

$\displaystyle =\frac1{x^{1/3}-y^{1/3}}\cdot\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}$

$\displaystyle =\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x-y}$

Note that the expression which I multiplied the numerator and denominator by is just the second half of the difference of cubes formula. - Jun 5th 2012, 04:53 AMpurplec16Re: Rationalize the denominator
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?

- Jun 5th 2012, 05:13 AMReckonerRe: Rationalize the denominator
- Jun 28th 2012, 09:14 AMpurplec16Re: Rationalize the denominator
Okay. Thanks for your help!