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Added answer to edited question

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edited Dec 25, 2016 at 16:14

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EDIT: For a multiple choice,

Select[
 {a^3 b^3, b^3 c^3, a^3 b^6, a^3 c^3, c^6},
 Assuming[
   {a == (a - b)^2 + b (a + 1),
    b == (b - c)^2 + c (b + 1),
    c == (c - a)^2 + a (c + 1)},
   Simplify[
    (a^6 - b^6)^2/(c^6 - 4 a^3 b^3) == #]] &]

(*  {c^6}  *)

EDIT: For a multiple choice,

Select[
 {a^3 b^3, b^3 c^3, a^3 b^6, a^3 c^3, c^6},
 Assuming[
   {a == (a - b)^2 + b (a + 1),
    b == (b - c)^2 + c (b + 1),
    c == (c - a)^2 + a (c + 1)},
   Simplify[
    (a^6 - b^6)^2/(c^6 - 4 a^3 b^3) == #]] &]

(*  {c^6}  *)

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created Dec 25, 2016 at 0:25

Bob Hanlon

162.7k
7
81
205

Solve[Eliminate[{A == (a^6 - b^6)^2/(c^6 - 4 a^3 b^3), 
    a == (a - b)^2 + b (a + 1), b == (b - c)^2 + c (b + 1), 
    c == (c - a)^2 + a (c + 1)}, {a, b, c}], A] // Simplify

(*  {{A -> -6 - 3*((1/2)*(123 - 
                   55*Sqrt[5]))^(1/3) - 
         3*((1/2)*(123 + 55*Sqrt[5]))^
             (1/3)}, 
   {A -> -6 + (3/2)*(1 + I*Sqrt[3])*
           ((1/2)*(123 - 55*Sqrt[5]))^
             (1/3) + (3 - 3*I*Sqrt[3])/
           (2^(2/3)*(123 - 55*Sqrt[5])^
                (1/3))}, 
   {A -> -6 + (3/2)*(1 - I*Sqrt[3])*
           ((1/2)*(123 - 55*Sqrt[5]))^
             (1/3) + (3 + 3*I*Sqrt[3])/
           (2^(2/3)*(123 - 55*Sqrt[5])^
                (1/3))}}  *)

% // N

(*  {{A -> -21.5225}, {A -> 1.76124 - 12.398 I}, {A -> 
   1.76124 + 12.398 I}}  *)