1. Dec 10, 2008

### Mike_Winegar

1. The problem statement, all variables and given/known data
-60/120 ^1/3

2. Relevant equations
I've gone through my book, and I'm supposed to find a number that when multiplied by a perfect cube I can bring it out and then cancel with the division. Try as I might with my calculator, I can't find one. Am I doing this wrong?

3. The attempt at a solution I broke the numerator and denominator down and I got 2 X 2 X 3 X 5 for 60, and 2 X 2 X 2 X 3 X 5 for 120. The only number I can bring out of the denominator is 2 X 2 X 2 which allows me to bring out 8. There is no number that I can bring out of the numerator. So basically, I got to 2 X 2 X 3 X 5 ^1/3 divided by 2(3 X 5)^1/3

Ahhh...I think that by talking this through, I figured it out. Can you let me know if I've done it right?

Can I cancel the 3 and 5 since they're both cube rooted? then that leaves me with 4^1/3 divided by 2?

Edit: Ehhh...Tried doing this on another problem, and ran into issues, so I'm guessing that I did this wrong.

Last edited: Dec 10, 2008
2. Dec 10, 2008

### Mike_Winegar

Forgive me, I haven't posted here in forever, and I've forgotten the code for entering actual symbols and such.

3. Dec 10, 2008

### HallsofIvy

Staff Emeritus
No need to worry about Latex code, that looks fine- but use parenthese. I think what you meant was (-60/120)^(1/3). Yes, factoring is a perfectly good way to find a root, although I think I would have been inclined to see that 120= 60*2 and stop there:
-60/120= -60/(2*60)= -1/2. Your problem reduces to (-1/2)^(1/3). That can be rewritten as -(1/2)^(1/3) but I can't see that you can do a lot after that. If you are allowed to use negative exponents you could write that as -1/2^(1/3)= -2^(-1/3).

4. Dec 10, 2008

### Mike_Winegar

Hmmm...No answer in the back of the book for the the (-60/120)^1/3. But the answer for (-60/180)^1/3 ends up being -[(9^1/3)/3]. Any ideas on how they got that?

5. Dec 10, 2008

### Mike_Winegar

Ok, I see. They got that answer by rationalizing the denominator. Thanks for you help, greatly appreciate it.