## kandsbbf Group Title How can you write the expression with rationalized denominator? one year ago one year ago

1. kandsbbf Group Title

@nooce

2. kandsbbf Group Title

@zepdrix

3. zepdrix Group Title

4. zepdrix Group Title

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

5. zepdrix Group Title

$\large \frac{1}{\sqrt2}$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 Group Title

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. kandsbbf Group Title

okay....

8. zepdrix Group Title

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

9. kandsbbf Group Title

ohhh okay (: i get ya

10. zepdrix Group Title

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

11. kandsbbf Group Title

i believe so lol

12. kandsbbf Group Title

now what?

13. zepdrix Group Title

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 Group Title

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

15. kandsbbf Group Title

so the bottom is 6 lol

16. zepdrix Group Title

Yah :3

17. kandsbbf Group Title

so now the top?

18. zepdrix Group Title

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

19. kandsbbf Group Title

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

20. zepdrix Group Title

oh ok that might help ^^

21. kandsbbf Group Title

okay hold on

22. kandsbbf Group Title

23. zepdrix Group Title

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 Group Title

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

25. kandsbbf Group Title

okay i see what ur sayying

26. zepdrix Group Title

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

27. zepdrix Group Title

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

28. zepdrix Group Title

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 Group Title

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

30. kandsbbf Group Title

soooo

31. zepdrix Group Title

Ah sorry lost my connection D:

32. kandsbbf Group Title

33. zepdrix Group Title

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

34. zepdrix Group Title

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

35. zepdrix Group Title

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

36. kandsbbf Group Title

algabra 2 :/

37. zepdrix Group Title

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

38. zepdrix Group Title

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

39. kandsbbf Group Title

okay thankyou(:could u help me with another?

40. kandsbbf Group Title

it's not like that one i promise lol

41. zepdrix Group Title

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 ^^