## anonymous 3 years ago How can you write the expression with rationalized denominator?

1. anonymous

@nooce

2. anonymous

@zepdrix

3. zepdrix

4. zepdrix

So we have an irrational number in the denominator. To fix that, let's ummm

5. zepdrix

$\large \frac{1}{\sqrt2}$Here's a quick example, when dealing with a SQUARE (2nd) root, we can multiply the top and bottom by $$\huge \frac{\sqrt2}{\sqrt2}$$ and it will end up changing the denominator to a rational number, namely, 2. But in this problem we have a CUBE (3rd) root, so we have to do something a lil bit fancier.

6. zepdrix

To get a 6 in the denominator we'll need to multiply the top and bottom by $$\huge \frac{6^{2/3}}{6^{2/3}}$$

7. anonymous

okay....

8. zepdrix

$\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}$Understand the fractional exponent notation ok? :)

9. anonymous

ohhh okay (: i get ya

10. zepdrix

$\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)$Understand how the bottom will simplify? c:

11. anonymous

i believe so lol

12. anonymous

now what?

13. zepdrix

When we MULTIPLY terms with similar bases, we ADD the exponents. So in the bottom we'll get,$\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}$

14. zepdrix

$\huge = \qquad 6^{3/3} \qquad = 6$

15. anonymous

so the bottom is 6 lol

16. zepdrix

Yah :3

17. anonymous

so now the top?

18. zepdrix

Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)

19. anonymous

i could give u the possible answers..... they all have 6 under them

20. zepdrix

oh ok that might help ^^

21. anonymous

okay hold on

22. anonymous

23. zepdrix

Oh ok I see what they want us to do. So now that we've taken care of the bottom, we'll write the $$\large 6^{2/3}$$ like this $$\large \sqrt[3]{6^2}$$ before we distribute it to the top.

24. zepdrix

$\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}$And I guess they want us to simplify a bit from here.

25. anonymous

okay i see what ur sayying

26. zepdrix

$\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}$

27. zepdrix

This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.

28. zepdrix

The following numbers are perfect cubes, 8, 27, 64, 125. Because if you take the cube root of any of those numbers, they'll give you a nice clean value.$\large 2^3=8, \qquad 3^3=27, \qquad ...$

29. zepdrix

It turns out that 108 is divisible by 27!$\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}$

30. anonymous

soooo

31. zepdrix

Ah sorry lost my connection D:

32. anonymous

33. zepdrix

27 is a perfect cube! So let's take the cube root of 27.

34. zepdrix

$\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3$

35. zepdrix

This is a really awful problem. Is this for algebra or something? D:

36. anonymous

algabra 2 :/

37. zepdrix

So it looks like it isssss probablyyyyyyy .... d. I'm like .. 65% sure :D lol

38. zepdrix

It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.

39. anonymous

okay thankyou(:could u help me with another?

40. anonymous

it's not like that one i promise lol

41. zepdrix

Close this thread. Start a new one with your new question. Type @zepdrix somewhere in the comments and I'll try to take a look if I have time ^^

Find more explanations on OpenStudy