Given that a, b, and c are all positive, and that ab = 21, ac = 18, and bc = 42, ï¬nd a + b + c.
269let,
ab=21........(1)
ac=18.........(2)
bc=42.........(3)
let us divide equation 1 and 2,we get
abac=2118
which gives,
bc=76
or, b=7c6
now,substituting this value in equation 3,we get,
[7c6]c=42
c2=42*67
c=6
therefore value of "a"
ac=18
a*6=18
a=3
and value of "b"
bc=42
b*6=42
b=7
therefore value of
a+b+c=3+7+6
=16
here is the answer .... :) :)
69I have another method to solve the sum.
ab = 21 = 3 * 7
bc = 42 = 2 * 3 * 7
ac = 18 = 2 * 3 *3
Multiplying the three equations,
we get a2b2c2 = 22 * 32 * 32 * 72
Therefore, abc = √22 * 32 * 32 * 72 = 2 * 3 * 3 * 7 = 126
Therefore a = abcbc = 12642 = 3
Therefore b = abcac = 12618 = 7
Therefore c = abcab = 12621 = 6
Therefore, a + b + c = 3 + 7 + 6 = 16
Answer) 16 [Proved]
69Manish Sir, is my process correct?
Rohit Ghosh absolutely wrongAngikar Ghosal how? which step is wrong? 
Manish Shankar yup its correct