SOLUTION: help 3sqrt-64 over x^3 please show work

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Question 35658: help
3sqrt-64 over x^3
please show work

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Do you mean "Cube root"??? This is NOT a square root!! Now, before we do any cube roots, you must know what the perfect cubes are!! That is:
2%5E3=8
3%5E3=+27
4%5E3=64 and
5%5E3=+125.

You MUST know these numbers before you do a CUBE ROOT problem!! Also, you must know that when you take a cube root of a variable raised to a power, you must DIVIDE the exponent by 3. If you don't know all of this, you should visit my own website by clicking on my tutor name "rapaljer", then look for Basic Algebra, then "Samples from Basic Algebra: One Step at a Time", then go to Chapter 5, and find "5.02 Cube Roots and More". That gives the entire explanation that you need, much more than I can give you right here.

Now, back to the problem!! It's really simple:
root%283%2C+%2864%2F%28x%5E3%29%29%29+
Remember cube root of 64 says, "What number can I cube to get 64?" The answer is 4.

And for the denominator, remember to divide the exponent by 3, which gives you x%5E1 or x.

Therefore, root%283%2C+%2864%2F%28x%5E3%29%29%29+=+4%2Fx.

R^2 at SCC