Radicals - Question is attached

Question

anonymous · Answer

attachment

radar · Answer

I can do that, I will do it in a series of steps.  The problem is equivalent to the following.  Do you see why?$$(2mn)^{1/3}\times(2mn)^{1/3}\times(2mn)^{1/3}$$

radar · Answer

$$(2mn)^{(1/3 + 1/3 +1/3})=(2mn)^{1}$$ or simply 2mn

radar · Answer

Earth to Hotwire........are you there?

anonymous · Answer

Ok, so the cube root thing cancels out the m^3n^3, right?

anonymous · Answer

Sorry, had to step away for a second.

radar · Answer

I think you may made a typo, but here are some facts, let me know if they don't make sense to you..$$\sqrt[3]{2mn}=(mn)^{1/3}   $$ Does that look familiar?

radar · Answer

Sorry I left out the 2, just imagine it is in both expressions (2mn)^1/3

anonymous · Answer

Right, that makes sense. But when I multiply the mn and the $$m^2n^2$$, I get $$m^3n^3$$, which is canceled out by the cube root, making the answer 2mn. Right?

radar · Answer

$$\sqrt[3]{4m ^{2}n ^{2}}=\sqrt[3](2)(2)(m)(m)(n)(n){}$$
Yes and also you have $$\sqrt[3]{2^{3}}$$which will give you 2.  thats it Hotwire, I think you are on it.

anonymous · Answer

Alright, Thanks a lot for the help. I actually get this now!

radar · Answer

I just converted them to fracional exponents and added.  Same thing.

radar · Answer

Good luck with them.