Re: How to instruct Math to take a certain (e.g. real)

*To*: mathgroup at smc.vnet.net*Subject*: [mg105051] Re: [mg105018] How to instruct Math to take a certain (e.g. real)*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 19 Nov 2009 05:23:19 -0500 (EST)*References*: <200911181157.GAA04186@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

Each cube root on the left has 3 possible values (except 0^(1/3), so the sum has nine possible values. Only three are zero. We can replace the terms with primary roots, after substituting y->0: Here's the method applied to each term separately: eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3); Cases[eq, Power[x_, Rational[1, n_Integer]] :> Root[#^n - x /. y -> 0 &, 1], Infinity] {-2, 2, 0} And here it is, applied to the equation: eq /. Power[x_, Rational[1, n_Integer]] :> Root[#^n - x /. y -> 0 &, 1] True Bobby On Wed, 18 Nov 2009 05:57:41 -0600, Alexei Boulbitch <Alexei.Boulbitch at iee.lu> wrote: > Dear Community, > > I came to a problem, that I cannot check the solution y=0 of equation > > In[345]:= eq = (-8 + y)^(1/3) + (8 + y)^(1/3) == y^(1/3); > eq /. y -> 0 > > Out[346]= False > > just because the expression > > In[347]:= eq[[1]] /. y -> 0 > > Out[347]= 2 + 2 (-1)^(1/3) > > Mathematica does not interpret as zero. And if I ask it to give a > numerical answer > In[348]:= 2 + 2 (-1.)^(1/3) > > Out[348]= 3.+ 1.73205 \[ImaginaryI] > It returns the complex root out of the three possible. > > My question is the following: > > 1) How should I instruct Mathematica to take a certain root that I want > of say, (-1)^(1/3)? > > 2) I think there is a general possibility instruct Mathematica that all > calculations should be done on reals only. Is it right? > > Thank you, Alexei > -- DrMajorBob at yahoo.com

**References**:**How to instruct Math to take a certain (e.g. real) type of results***From:*Alexei Boulbitch <Alexei.Boulbitch@iee.lu>