how do i start

**kandela573** · July 15th 2009, 02:22 PM

I need to rationalize: 9 cuberoot of 2 over cuderooot of 5 - cuberoot of 2

**Plato** · July 15th 2009, 02:29 PM

Recall that $(a^3 -b^3)=(a-b)(a^2+ab+b^2)$ .
$\frac{{\sqrt[3]{9}}}{{\sqrt[3]{5} - \sqrt[3]{2}}}\left( {\frac{{\sqrt[3]{{25}} + \sqrt[3]{{10}} + \sqrt[3]{4}}}{{\sqrt[3]{{25}} + \sqrt[3]{{10}} + \sqrt[3]{4}}}} \right)=?$

**kandela573** · July 15th 2009, 02:38 PM

okay but in order for it to be simplified i can't have a radicand in the denominator so do i mulitply everything by the third power?

**VonNemo19** · July 15th 2009, 02:59 PM

Originally Posted by kandela573

okay but in order for it to be simplified i can't have a radicand in the denominator so do i mulitply everything by the third power?

Follow through with the multiplication that Plato has indicated and see what happens.

**Plato** · July 15th 2009, 03:41 PM

Originally Posted by kandela573

okay but in order for it to be simplified i can't have a radicand in the denominator so do i mulitply everything by the third power?

Can you do this, $\left( {\sqrt[3]{5} - \sqrt[3]{2}} \right)\left( {\sqrt[3]{{25}} + \sqrt[3]{{10}} + \sqrt[3]{4}} \right)=?$

If not, why were you asked to do this problem?

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