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
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