Question

if a²=b+c,b²=c+a and c²=a+b,then the value of (1/1+a)+(1/1+b)+(1/1+c) is

Answer 1

a² = b + c
b² = c + a
c² = a + b

a² - b² = b - a      =>  (a - b) (a + b) = b - a
               =>  a + b  = -1    or  a = b

1)  a + b ≠ - 1      as  c²  is not negative, if c is a real number.
         a = b  =>  c² = 2 a
     similarly, b = c    =>  a² = 2 b            (as  b+c ≠ -1, as a² is non negative)
       and  c = a =>  b² = 2 a

   so we get a = b = c = 2

$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\\\\=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

Answer 2

(1/1+a)+(1/1+b) +(1/1+c)  take LCm 1/1+a+1/1+b+1/+1/1+c = {(b+1)(c+1)+(c+1)(a+1)+(a+1)(b+1)}/(a+1)(b+1)(c+1)
solve the equation 
{bc+b+c+1+ac+c+a+1+ab+a+b+1}/(ab+b+a+1)(c+1)
(a+a+b+b+c+c+ab+bc+ca) /(ab+b+1+a)(c+1)
we know that 
a²=b+c  , b²=c+a , c²=a+b 
{(b+c)+(b+c) + (a+a)}/(ab+1+(b+c))(c+1)
{a²+a²+2a }/{abc+c+c(b+c)+ab+1+(b+c)}
2a(a+1)/abc+c+c(a²)+ab+1+(a²)
2a(a+1)/abc+c+a²c+ab+1+a²
2a(a+1)/abc+ c(a²+1) +a(b+a)+1
2a(a+1)/abc+c²(a+1)+ac²
2a(a+1)/abc+ac²+c²+ac²
2a(a+1)/abc+c²(a+1+a)
2a(a+1)/abc+c²(2a+1)
2a(a+1)/c(ab+c(2a+1)
2a(a+1)/c(ab+2ac+c)
2a(a+1)/c(ab+ac+ac+c)
2a(a+1)/c(a(b+c)+c(a+1)
2a(a+1)/c(a(a²)+c(a+1)
2a(a+1)/c(a³+ac+c) by solving this we get 
2a(a+1)/2a(a+1) = 1