Rationalize the denominator

**purplec16** · June 3rd 2012, 08:17 PM

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

**purplec16** · June 3rd 2012, 08:18 PM

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

**Reckoner** · June 3rd 2012, 08:58 PM

$\frac1{x^{1/3}-y^{1/3}}$

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

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

Note that the expression which I multiplied the numerator and denominator by is just the second half of the difference of cubes formula.

**purplec16** · June 5th 2012, 04:53 AM

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?

**Reckoner** · June 5th 2012, 05:13 AM

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?

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

Distributing, we have

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

Now, combining like terms leaves

$=x-y$ .

**purplec16** · June 28th 2012, 09:14 AM

Okay. Thanks for your help!

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