What is the maximum and minimum sum for $|a-b|+|b-c|+|c-a|$ with given relation?

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Discussion Overview

The discussion centers on finding the maximum and minimum values of the expression $|a-b|+|b-c|+|c-a|$ under the constraint given by the equation $(a-b)^3+(b-c)^3+(c-a)^3 = 60$, involving integer variables $a$, $b$, and $c$. The scope includes mathematical reasoning and exploration of the relationship between the variables.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant initiates the discussion by posing the problem of maximizing and minimizing the sum $|a-b|+|b-c|+|c-a|$ given the cubic relation.
  • The same participant reiterates the problem statement, emphasizing the constraint involving the cubes of the differences.
  • A third participant offers a brief acknowledgment of the initial post, indicating engagement with the topic.

Areas of Agreement / Disagreement

The discussion remains unresolved, with no consensus reached on the maximum and minimum values of the expression.

Contextual Notes

The discussion does not explore specific methods or solutions to the problem, and no assumptions or definitions are clarified regarding the variables or the nature of the cubic relation.

lfdahl
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Find the maximum and minimum of the sum $|a-b|+|b-c|+|c-a|$, if the integer numbers $a,b$ and $c$ satisfy the following relation:

\[ (a-b)^3+(b-c)^3+(c-a)^3 = 60 \]
 
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lfdahl said:
Find the maximum and minimum of the sum $|a-b|+|b-c|+|c-a|$, if the integer numbers $a,b$ and $c$ satisfy the following relation:

\[ (a-b)^3+(b-c)^3+(c-a)^3 = 60 \]

using $x + y +z = 0 => x^3+y^3 + z^3 = 3xyz$ we get letting x = a-b , y = b-c , z = c - a = -(x+y)
$3(a-b)(b-c)(c-a) = 60$ or $(a-b)(b-c)(c-a) = 20$
or $xy(x+y) = - 20$
taking factors of -20 (product of 3 numbes) such that product is 20 we get 3 triplets ( -1,-4,-5), (-1,5,4), (-4,5,1) and in all
cases $|x| + |y| + |z|$ or $|a-b| + |b-c| + |c-a| = 10$

Needless to say that both maximum and minimum are 10
 
kaliprasad said:
using $x + y +z = 0 => x^3+y^3 + z^3 = 3xyz$ we get letting x = a-b , y = b-c , z = c - a = -(x+y)
$3(a-b)(b-c)(c-a) = 60$ or $(a-b)(b-c)(c-a) = 20$
or $xy(x+y) = - 20$
taking factors of -20 (product of 3 numbes) such that product is 20 we get 3 triplets ( -1,-4,-5), (-1,5,4), (-4,5,1) and in all
cases $|x| + |y| + |z|$ or $|a-b| + |b-c| + |c-a| = 10$

Needless to say that both maximum and minimum are 10

Later I realized that there is only one solution (a-b=5, b-c = -4, c-a = -1) or a permutation of the same permutation counting the same
'
 
kaliprasad said:
Later I realized that there is only one solution (a-b=5, b-c = -4, c-a = -1) or a permutation of the same permutation counting the same
'

Well done!

Yes, the only solution is the set $(5,-4,-1)$ and its permutations.
 

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