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Derive a formula (Posted on 2023-07-27) Difficulty: 2 of 5
Given
a+b+c=0 and abc=4

Evaluate S=a^3+b^3+c^3

See The Solution Submitted by Ady TZIDON    
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Solution computer solution | Comment 2 of 3 |
This works:

If, say, b and c each are negative and a = -(b + c), and in fact we make b = c, let b and c be -x and a be 2*x. This assures a+b+c = 0.

2*x*x*x = 4

x^3 = 2

a=2*2^(1/3)
b=c=-2^(1/3)

S = a^3+b^3+c^3 = 8*2 - 2 - 2 = 16 - 4 = 12


But I don't know how to show some other combination might not come out differently.

However, Wolfram Alpha solves it as

  a not equal zero

  b = -(a^(3/2) + sqrt(a^3 - 16))/(2*sqrt(a));
  c = sqrt(a^3 - 16)/(2*sqrt(a)) - a/2;

so

clc
for a=vpa([-10:-1 1:10])
  b = -(a^(3/2) + sqrt(a^3 - 16))/(2*sqrt(a));
  c = sqrt(a^3 - 16)/(2*sqrt(a)) - a/2;
  disp([a b c a^3+b^3+c^3])
  disp(' ')
end

finds

 a               b                                      c                     a^3+b^3+c^3

[-10.0, -0.039841267341661029428932185154232, 10.039841267341661029428932185154, 12.0]
 
[-9.0, -0.04911468798539354262640114557686, 9.0491146879853935426264011455769, 12.0]
 
[-8.0, -0.06201920231798018022994178413302, 8.062019202317980180229941784133, 12.0]
 
[-7.0, -0.08070224556979485084870026652759, 7.0807022455697948508487002665276, 12.0]
 
[-6.0, -0.10912635102960501518482518077352, 6.1091263510296050151848251807735, 12.0]
 
[-5.0, -0.15518360947035073399678594261552, 5.1551836094703507339967859426155, 12.0]
 
[-4.0, -0.23606797749978969640917366873128, 4.2360679774997896964091736687313, 12.0]
 
[-3.0, -0.39296944860009120363077932799745, 3.3929694486000912036307793279975, 12.0]
 
[-2.0, -0.73205080756887729352744634150587, 2.7320508075688772935274463415059, 12.0]
 
[-1.0, -1.561552812808830274910704927987, 2.561552812808830274910704927987, 12.0]
 
[1.0, - 0.5 - 1.9364916731037084425896326998912i, - 0.5 + 1.9364916731037084425896326998912i, 12.0]
 
[2.0, - 1.0 - 1.0i,                         - 1.0 + 1.0i,                     12.0]
 
[3.0, -2.4574271077563381099751019113698, -0.54257289224366189002489808863018, 12.0]
 
[4.0, -3.7320508075688772935274463415059, -0.26794919243112270647255365849413, 12.0]
 
[5.0, -4.8345235059857504034915031069726, -0.16547649401424959650849689302735, 12.0]
 
[6.0, -5.8867513459481288225457439025098, -0.11324865405187117745425609749021, 12.0]
 
[7.0, -6.9173924896873388797273933338048, -0.08260751031266112027260666619518, 12.0]
 
[8.0, -7.937003937005905509842517224406, -0.062996062994094490157482775593996, 12.0]
 
[9.0, -8.9503433076062295573580612844326, -0.049656692393770442641938715567387, 12.0]
 
[10.0, -9.9598387070548977681702045553081, -0.040161292945102231829795444691874, 12.0]
 
Note for small positive a (i.e., here 1 and 2), that b and c are complex but the sum of the cubes is still 12.

The general answer is S = 12.

Of course the values of a, b and c can be interchanged.



  Posted by Charlie on 2023-07-27 09:18:06
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