radical equations

I suppose this is what you mean?
\({^{5}}{sqrt{-32{x^{10}}{y^{5}}}}\)

To solve this, we should rewrite everything in exponentiated form:
\({^{5}}{sqrt{-{2^{5}}{x^{10}}{y^{5}}}}\)

Now, we divide each of the exponents by the index of the radical, 5:
-25: 5/5 = 1 R 0
x10: 10/5 = 2 R 0
y5: 5/5 = 1 R 0

The answers of the division are the exponents for each base that will come out of the radical, and the remainders are the exponents that will remain under the radical.  Since we don't have any remainders, the radical will be eliminated:

Answer:
-21x2y1 = -2x2y

Quick side note, had the index of the radical been even instead of odd, there would have been no real answer solution since you cannot take an even index radical of a negative number and get a real solution.

Also remember that, when you have numbers multiplied or divided with each other (for example a*b or a/b) you can do the following:
\({sqrt{ab}\) = \({sqrt{a}\)\({sqrt{b}\)

So, in your example, you could also do:
\({^{5}}{sqrt{-32{x^{10}}{y^{5}}}}\)=\({^{5}}{sqrt{-32}}\) \({^{5}}{sqrt{x^{10}}}\) \({^{5}}{sqrt{y^{5}}}\) and then do the same thing as Doctor-2-B separately for each member.

Just be careful, the same does not apply when you have subtractions or abstractions.
(for example, \({sqrt{a+b}\) is not equal with \({sqrt{a}\)+\({sqrt{b}\), and therefore, in such cases, you can't simplify using that method.

Basically my advice isn't very good, because you could end up with negative square roots etc.
Just stick with what Doctor-2-B said, and remember you can do the save with divisions too.