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Simplify question

• To: mathgroup at smc.vnet.net
• Subject: [mg71655] Simplify question
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Sun, 26 Nov 2006 03:48:36 -0500 (EST)

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

• Follow-Ups:
  • Re: Simplify question
    • From: Daniel Lichtblau <danl@wolfram.com>
  • Re: Simplify question
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>