a²= b+cb²=c+ac²= a+bThen 1/1+a + 1/1+b + 1/1+c=?

Solution.

Given : a²= b+c – (1)
b²= c+a – (2)
c²= a+b – (3)

Now, subtract (1) – (2)
a²– b²= (b+c) – (c+a)
(a+b)(a-b) = – (a-b) – (4)
Fom above eqn we have two results
a+b = – 1or a-b=0
But a+b = – 1 is not possible because from eqn (3), c²or a+b must be a positive integer, hence a=b holds true
Now from eqn (3), c²= 2a
In a similar way we can solve eqn (2) and (3) and we get,
b=c or a²= 2b
and
c=a or b²= 2a
After solving this we get, a=b=c=2.
Hence the value of expression,
1/(1+a) + 1/(1+b) + 1/(1+c) = 1/3 + 1/3 + 1/3 = 1. Ans

a=( b+c)/a 1+a= (a+b+c)/a
b= (a+c)/b 1+b= (a+b+c)/b
c= (a+b)/c 1+c= (a+b+c)/c
1/(1+a) + 1/(1+b) + 1/(1+c)= (a+b+c)/(a+b+c) = 1
so the answer will be 1.