# how do i start

• Jul 15th 2009, 03:22 PM
kandela573
how do i start
I need to rationalize: 9 cuberoot of 2 over cuderooot of 5 - cuberoot of 2
• Jul 15th 2009, 03:29 PM
Plato
Recall that $(a^3 -b^3)=(a-b)(a^2+ab+b^2)$.
$\frac{{\sqrt[3]{9}}}{{\sqrt[3]{5} - \sqrt[3]{2}}}\left( {\frac{{\sqrt[3]{{25}} + \sqrt[3]{{10}} + \sqrt[3]{4}}}{{\sqrt[3]{{25}} + \sqrt[3]{{10}} + \sqrt[3]{4}}}} \right)=?$
• Jul 15th 2009, 03:38 PM
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?(Thinking)
• Jul 15th 2009, 03:59 PM
VonNemo19
Quote:

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?(Thinking)

Follow through with the multiplication that Plato has indicated and see what happens.
• Jul 15th 2009, 04:41 PM
Plato
Quote:

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?(Thinking)

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?