## anonymous one year ago I need help with this problem trying to figure out how to do it, plz help. I will give medal.

1. anonymous

2. anonymous

help plz

3. anonymous

@phi

4. phi

if you have a "cube root" then if you multiply it by itself *three times* , you "get rid" of the the radical when they say "rationalized denominator" it means get rid of the radical in the bottom part of the fraction so to do that, multiply the bottom by $$\sqrt[3]{6} \sqrt[3]{6}$$ and to keep things equal , you have to multiply the top by the same thing you get $\frac{\sqrt[3]{2} \sqrt[3]{6} \sqrt[3]{6} }{\sqrt[3]{6} \sqrt[3]{6} \sqrt[3]{6} }$

5. phi

by definition, the bottom becomes 6 (the cube root multiplied three times undoes the radical)

6. anonymous

Why would you multiply the numerator and the denominator twice by the denominator

7. anonymous

oh ok

8. anonymous

because it is cubed

9. phi

in the top, all three are cube roots so you can combine the top into $\sqrt[3]{2\cdot 6 \cdot 6}$ I would factor the 6 into 2*3 so we can write the top as $\sqrt[3]{2\cdot 2 \cdot 3 \cdot 2 \cdot 3}$ or reordering to make it clearer $\sqrt[3]{2\cdot 2 \cdot 2 \cdot 3 \cdot 3}$ now if we find *three of the same term* inside a cube root. we can "pull them out" and replace them with 1 term on the outside $2 \sqrt[3]{3\cdot 3}$ so far we have $\frac{2 \sqrt[3]{3\cdot 3}}{6}$

10. anonymous

nvmd

11. anonymous

and then multiply the 3's on the inside of the radical

12. phi

can you finish? we can't do much with the 3's inside (we would need 3 of them to simplify) so I would make them 9 and of course we can simplify 2/6 on the outside

13. anonymous

so it would be A.

14. phi

yes

15. anonymous

ok I understand how to do it now thankyou very much

16. phi

yw