SOLUTION: I asked this question a few minutes ago and I wrote it wrong. Rapaljer was kind enough to try to answer it but because of my mistake his answer doesn't help me:) The question is:

Algebra ->  Square-cubic-other-roots -> SOLUTION: I asked this question a few minutes ago and I wrote it wrong. Rapaljer was kind enough to try to answer it but because of my mistake his answer doesn't help me:) The question is:      Log On


   

Question 50653: I asked this question a few minutes ago and I wrote it wrong. Rapaljer was kind enough to try to answer it but because of my mistake his answer doesn't help me:)
The question is:
Fourth root of quantity: 81*x^6*y^8
Thank you:)
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Okay, if it is a fourth root, then you need to recognize that 81 is a perfect 4th power, that is 3%5E4+=+81 so root%284%2C+81%29=3. Also, you need to know that if you have a variable raised to a power, then you divide the exponents by 4. The problem here is that the exponent of x is 6 which is NOT divisible by 4. So you have to break it down and write x%5E6=+x%5E4%2Ax%5E2.

root%284%2C+81x%5E6%2Ay%5E8%29+
root%284%2C+81x%5E4%2Ay%5E8%29+%2Aroot%284%2C+x%5E2%29
+3xy%5E2%2A+root%284%2C+x%5E2%29

Now, there is an additional problem, that the root%284%2Cx%5E2%29 reduces to sqrt%28x%29+. This is called reducing the order of a radical, which to my way of thinking is a College Algebra skills. Have you had anything like that??

So the last step becomes this:
+3xy%5E2%2A+root%284%2C+x%5E2%29
+3xy%5E2%2A+sqrt%28+x%29


R^2 at SCC