# Find a,b,c

#### Albert

##### Well-known member
$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$find :\,\, a,b,c$

##### Well-known member
$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$find :\,\, a,b,c$
a = 3, b= 1, c= 6

for solution
it is not elegent but effecitive

from the ( a − b)(b + c)(c − a ) = 42
as 7 devides 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 then they are seen to satisfy other 2 equations

#### Opalg

##### MHB Oldtimer
Staff member
$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$find :\,\, a,b,c$
My solution, like kaliprasad's, involved a bit of guesswork. Add the first and second equations to get $$-2(a-b)^2c = -48.$$ Add the second and third equations to get $$-2(c-a)^2b = -18.$$ Add the first and third equations to get $$-2(b-c)^2a = -150.$$ You can write those equations as $$(a-b)^2c = 24 = 2^2*6,$$ $$(c-a)^2b = 9 = 3^2*1,$$ $$(b-c)^2a = 75 = 5^2*3.$$ If you then make the assumption that the squared terms on the left side of those equations correspond to the squared terms on the right side, and the nonsquared terms likewise correspond, then you read off the solution $a=3,\,b=1,\,c=6,$ which turns out to be correct.

My solution, like kaliprasad's, involved a bit of guesswork. Add the first and second equations to get $$-2(a-b)^2c = -48.$$ Add the second and third equations to get $$-2(c-a)^2b = -18.$$ Add the first and third equations to get $$-2(b-c)^2a = -150.$$ You can write those equations as $$(a-b)^2c = 24 = 2^2*6,$$ $$(c-a)^2b = 9 = 3^2*1,$$ $$(b-c)^2a = 75 = 5^2*3.$$ If you then make the assumption that the squared terms on the left side of those equations correspond to the squared terms on the right side, and the nonsquared terms likewise correspond, then you read off the solution $a=3,\,b=1,\,c=6,$ which turns out to be correct.