Problem to evaluate cube root of a negative cube nember where a real value is expected
- To: mathgroup at smc.vnet.net
- Subject: [mg28042] Problem to evaluate cube root of a negative cube nember where a real value is expected
- From: "Gary" <garylga at magix.com.sg>
- Date: Thu, 29 Mar 2001 03:24:09 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hi, Does anyone know a simplier way to simplify (-8)^(1/3)=(-2) other than what I did below because no complex answer is expected in the solution. In[161]:= p=(-8)^(1/3) q=Abs[p] Level[p,3] r=Extract[Level[p,3],2] (* if p is real, then p should be as below *) q*r Out[161]= \!\(2\ \((\(-1\))\)\^\(1/3\)\) Out[162]= 2 Out[163]= \!\({2, \(-1\), 1\/3, \((\(-1\))\)\^\(1/3\)}\) Out[164]= -1 Out[165]= -2 ______________________________________ Gary Lee Guanan (garylga at magix.com.sg) Director - Business Development IQExplorers.com Pte Ltd Tel 874-1345/6 ========================C/o Address============================ Incubation Centre, School of Computing(SoC), NUS. S15 #01-09, 1 Science Drive 2 (along Lower Kent Ridge Road), S117543 ===============================================================