Re: Simplify question

*To*: mathgroup at smc.vnet.net*Subject*: [mg71725] Re: [mg71655] Simplify question*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 27 Nov 2006 04:05:09 -0500 (EST)*References*: <200611260848.DAA14451@smc.vnet.net>

On 26 Nov 2006, at 17:48, dimitris wrote: > The following list of expressions was obtained by following the steps > of Tartaglia's solution of the cubic equation with Mathematica. > > lstcub = {{((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)* > (R - > Sqrt[Q^3 + R^2])^(1/3), > -(Q/(R - Sqrt[Q^3 + R^2])^(1/3)) + (R - Sqrt[Q^3 + R^2])^(1/3), > -(((-1)^(2/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3)) - > (-1)^(1/3)*(R - Sqrt[Q^3 + R^2])^(1/3)}, {-(Q/(R + Sqrt[Q^3 + > R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3), > ((-1)^(1/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R + > Sqrt[Q^3 + R^2])^(1/3), > -(((-1)^(2/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3)) - (-1)^(1/3)*(R + > Sqrt[Q^3 + R^2])^(1/3)}}; > > TableForm[%]//TraditionalForm > > Although the solution of the reduced cubic equation can be obtained > making a tricky observation > > (see e.g. Leonard E. (Leonard Eugene) Dickson b. 1874. (page 32) , > Elementary theory of equations. 1914. > available online at > http://mathbooks.library.cornell.edu:8085/Dienst? > verb=Display&protocol=CGM&ver=1.0&identifier=cul.math/01460001" > > target=_blank >> http://mathbooks.library.cornell.edu:8085/Dienst? >> verb=Display&protocol=CGM&ver=1.0&identifier=cul.math/01460001) > > so what I ask don't play any important role in the solution procedure, > I wonder if Mathematica can verify the equalities > > MapThread[Equal, lstcub] > {((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R - Sqrt[Q^3 > + R^2])^(1/3) == > -(Q/(R + Sqrt[Q^3 + R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3), > -(Q/(R - Sqrt[Q^3 + R^2])^(1/3)) + (R - Sqrt[Q^3 + R^2])^(1/3) == > ((-1)^(1/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3) + > (-1)^(2/3)*(R + Sqrt[Q^3 + R^2])^(1/3), -(((-1)^(2/3)*Q)/(R - > Sqrt[Q^3 + R^2])^(1/3)) - > (-1)^(1/3)*(R - Sqrt[Q^3 + R^2])^(1/3) == -(((-1)^(2/3)*Q)/(R + > Sqrt[Q^3 + R^2])^(1/3)) - > (-1)^(1/3)*(R + Sqrt[Q^3 + R^2])^(1/3)} > > which can be justified in view of the results > > Table[lstcub /. {R -> Random[], Q -> Random[]}, {5}] // Chop > > Thanks a lot for any response. > > Dimitris > It clearly can't because: In[1]:= FindInstance[((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R - Sqrt[Q^3 + R^2])^(1/3) != -(Q/(R + Sqrt[Q^3 + R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3), {R, Q}] Out[1]= {{R -> -7 + (15*I)/2, Q -> 18/5 + (34*I)/5}} etc, Andrzej Kozlowski Tokyo, Japan

**References**:**Simplify question***From:*"dimitris" <dimmechan@yahoo.com>