What is the 6th root of 256? ...left in radical notation

Question

anonymous · Answer

6th root of 256 = 2.5198421

anonymous · Answer

thanks for responding, but I was looking for the answer in radical notation

accessdenied · Answer

What I would do is break it into prime factorization and try to take out "perfect 6th root" factors.

pfenn1 · Answer

$$\sqrt[6]{256}$$

anonymous · Answer

ok, 
$$\sqrt[6]{256} = \sqrt[6]{64\times4} = 2\sqrt[6]{4}  ?$$

accessdenied · Answer

Yeah, that's what I get as well.  :)

anonymous · Answer

Hmm...the answer Im suppposed to get is $$2\sqrt[3]{2}$$

accessdenied · Answer

Ohh, I see that we should also use the fact that "(2^2)^(1/6) = 2^(2/6) = 2^(1/3)"

accessdenied · Answer

or in radical form...
$ \sqrt[6]{2^2} = (\sqrt[6]{2})^2 = \sqrt[3]{2} $

anonymous · Answer

wow, thanks for your response...I guess when they mean "write the expression in simplest form"  they really mean simplest form!

anonymous · Answer

I don't always "see" the simplest form in these types of problems!

accessdenied · Answer

You're welcome.  :)

I was thinking about if we needed the radical reduced as well in this case since I've never ran across that situation in my class.  :P

anonymous · Answer

Do you think since
$$2\sqrt[6]{4} = 2\sqrt[3]{2}$$
it is safe to assume if there was anothther perfect sqaure inside the square root sign, where the 4 is, then I would have to simplify further?

accessdenied · Answer

Yeah.

  I think it's easiest to understand the simplification when you express the radical as a rational power, a fraction.
  If we have some expression to a power inside the radical, then we can consider reducing the power and index as if they were a fraction.
   $\sqrt[6]{25} = \sqrt[6]{5^2} = 5^{2/6} = 5^{1/3} $

  A situation I'm unsure of, however, is if you have an extra factor there, though, like 3*25.  It doesn't really seem simpler to break it into TWO radicals to me.  That may be something to ask the teacher about.  :P

anonymous · Answer

thanks for your insight!  It has helped me see better when to simplify further

accessdenied · Answer

You're welcome!  Again, these are some really rare cases to deal with, from my experience.  :P

anonymous · Answer

I need to post another question, I have to close this one!