SOLUTION: When I try to graph or solve for (x)^(1/3) or other odd roots, I get a complex number when x is negative. I know that (-2)^3 = -8, so why isn't (-8)^1/3 =-2? I am trying to learn

Algebra ->  Radicals -> SOLUTION: When I try to graph or solve for (x)^(1/3) or other odd roots, I get a complex number when x is negative. I know that (-2)^3 = -8, so why isn't (-8)^1/3 =-2? I am trying to learn      Log On


   

Question 30507: When I try to graph or solve for (x)^(1/3) or other odd roots, I get a complex number when x is negative. I know that (-2)^3 = -8, so why isn't (-8)^1/3 =-2? I am trying to learn how to do the basics on a TI-89 T.
Thanks, Bruce
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
But it is!
%28-8%29%5E%281%2F3%29+=+-2 and if you were to cube both sides:
%28%28-8%29%5E%281%2F3%29%29%5E3+=+%28-2%29%5E3 = -8
On both the TI-86 and the TI-89 Titanium, I get:
%28-8%29%5E%281%2F3%29+=+-2