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

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The maximum and minimum values of the expression $|a-b|+|b-c|+|c-a|$ under the constraint $(a-b)^3+(b-c)^3+(c-a)^3 = 60$ have been determined. The maximum sum occurs when the differences between the integers $a$, $b$, and $c$ are maximized, leading to a value of 12. Conversely, the minimum sum is achieved when the integers are as close as possible, resulting in a value of 6. These conclusions are derived from analyzing the properties of cubic equations and absolute values.

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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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