Question

If a,b and c are positive real numbers such that
a(b+c)= 152,
b(c+a) =162 and c(a+b) =110 then abc =?

Answer 1

given,
a(b+c)=152
b(c+a)=162
c(a+b)=110
multiply all three equation ,
abc(a+b)(b+c)(c+a)=152 x 162 x 110--------(1)
now we have required
(a+b)(b+c)(c+a)

now we use AM≥GM rule
(a+b), (b+c) ,(c+a) are positive integer
so,
{(a+b)+(b+c)+(c+a)}/3≥{(a+b)(b+c)(c+a)}^1/3

2(a+b+c)=3{(a+b)(b+c)(c+a)}^1/3

now take cube both side ,
8(a+b+c)^3=27(a+b)(b+c)(c+a)-----------(2)

but a, b, c are positive
this is also satisfied rule of AM≥GM
now,
(a+b+c)=3(abc)^1/3
take cube both sides
(a+b+c)^3=27abc
put this equation (2)
8 x 27abc =27(a+b)(b+c)(c+a)

8abc=(a+b)(b+c)(c+a)---------(3)
put equation (3) in (1)
abc x 8abc=162 x 152 x110

(abc)^2=162 x 152 x 110/8
(abc)^2=81 x 76 x 55

abc=√{81 x 76 x 55}
=581.8

Answer 2

I think the third equation is typed wrong...
a(b+c) = 152       -- (1) 
b(c+a) = 162       -- (2)
c(a+b) = 170   --- (3)
Add all the three equations:    2 (ab+bc+ ca) = 152+162+110 = 484
So   ab + bc + ca = 242    -- (10)

(10) - (1) :  bc = 90
(10) - (2) :  ac = 80
(10) - (3) :  ab = 72

Multiply these three:  (abc)² = 90 *80*72
      abc = 720
That is the option  (d)

===================  with old given data:
a(b+c) = 152       -- (1) 
b(c+a) = 162       -- (2)
c(a+b) = 110   --- (3)

(2) - (1):  c (b-a) = 10  -- (4)
(3) / (4) :   (a+b) / (b-a) = 11
       =>  12 a = 10 b       =>    b = 1.2 a        -- (5)

Substitute (5) in (3):   c * 2.2 a = 110        =>   c a = 50   --(6)
Substitute (6) in (1):   a b = 152 -50 = 102         -- (7)
Substitute (7) in (2):   b c = 60    ---(8)

(6) * (7) * (8):  ca *ab * bc = (abc)² = 50*102*60 
=>  abc = 60 * √85

==================    Another way:

Add all the three equations:    2 (ab+bc+ ca) = 152+162+110 = 424
So   ab + bc + ca = 212    -- (10)

(10) - (1) :  bc = 60
(10) - (2) :  ac = 50
(10) - (3) :  ab = 102

So  abc)² = 60 *50*102
      abc = 60* √85