Q3/083) solve

Given a,b,c integers solve for a,b,c

(a−b)(b−c)(c+a)=−90 ..1
(a−b)(b+c)(c−a)=42.. 2
(a+b)(b−c)(c−a)=−60 ..3

  

Solution

from the (3)

as 7 divides RHS and not the other 2 equations we have

b+ c = 7 ( as b and c > 0)

so we get (a-b)(c-a) = 6

so combinations a-b = 6 , c- a = 1
a- b= 3 , c - a = 2
a - b = 2 c - a = 3
a- b = 1 , c- a = 6
and (-6, -1),(-3,-2), (-2,-3) and (-1,6) each can be tried with b+ c to give
a= 3, b = 1 and c = 6 then they are seen to satisfy other 2 equations