# Rationalize the denominator

• Jun 3rd 2012, 08:17 PM
purplec16
Rationalize the denominator
[\MATH]\frac{1}{((x)^(1/3)-(y)^(1/3))}[/MATH]
• Jun 3rd 2012, 08:18 PM
purplec16
Re: Rationalize the denominator
\frac{1}{((x)^(1/3)-(y)^(1/3))}
• Jun 3rd 2012, 08:58 PM
Reckoner
Re: 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 AM
purplec16
Re: 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 AM
Reckoner
Re: Rationalize the denominator
Quote:

Originally Posted by purplec16
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?

$\displaystyle \left(x^{1/3}-y^{1/3}\right)\left(x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}\right)$

Distributing, we have

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

Now, combining like terms leaves

$\displaystyle =x-y$.
• Jun 28th 2012, 09:14 AM
purplec16
Re: Rationalize the denominator