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

[lfdahl]

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:

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

 

[kaliprasad]

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:

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

Spoiler

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]

Spoiler

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

Spoiler

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
'

 

[lfdahl]

Spoiler

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!

Spoiler

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