Solving a System of Equations in N to Find a,b,c

[Albert1]

$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$ find :\,\, a,b,c$

 

[kaliprasad]

$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$ 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]

$a,b,c,\in N$
$( a −b)(b − c )(c + a ) = −90$
$( a − b)(b + c)(c − a ) = 42$
$( a + b)(b − c )(c − a ) = −60$
$ find :\,\, a,b,c$

My solution, like http://www.mathhelpboards.com/members/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.

 

[kaliprasad]

My solution, like http://www.mathhelpboards.com/members/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.

we can combine (from above)
(a−b)^2c=24=2^2∗6
and b+ c = 7 ( from by solution) and confirm c = 6
we could have (a-b) = 1 and c = 24 from the above and by trial and error remove it.
and combining the two we get c =6 ad confirm and then b= 1 and a = 3

combining the above 2 get less range to guess

 

[CellCoree]

To solve this system of equations in N, we will use a combination of algebraic manipulation and trial and error. First, we can simplify the equations by expanding the brackets and combining like terms. This will give us:

$a^2b - ab^2 + bc^2 - c^2a + a^2c + bc^2 + ac^2 + a^2b = -90$
$a^2b - ab^2 + bc^2 - c^2a + a^2c + b^2c - ac^2 - a^2b = 42$
$a^2b - ab^2 + b^2c - c^2a + a^2c - ac^2 - a^2b + b^2c - c^2a + ac^2 - a^2b = -60$

Next, we can combine the equations to eliminate variables. By adding the first and second equations, we can eliminate the term $a^2b$, and by subtracting the second equation from the third, we can eliminate the term $ac^2$. This will give us:

$2bc^2 - 2a^2b + 2a^2c + 2b^2c - 2c^2a - 2a^2b = -48$

From here, we can factor out the common terms and divide both sides by 2 to get:

$bc^2 - a^2b + a^2c + b^2c - c^2a - a^2b = -24$

We can then rearrange this equation to get:

$a^2b - a^2c + b^2c - c^2a = -24$

Now, we can use trial and error to find possible solutions for $a, b,$ and $c$. We know that $a, b,$ and $c$ must be positive integers, so we can start by trying different values for $a$, $b$, and $c$ such as 1, 2, and 3. Plugging these values into the equation, we can see that they do not satisfy the equation. We can continue trying different values until we find a combination that satisfies the equation.

After trying a few different values, we can find that $a=3$, $b=2$, and $c=4$ satisfies the equation