Given, `a^(2) =b+c, b^(2)=c + a " & " c^(2) = a + b` or `(a^(2))/(b+c) = (b^(2))/(c+a) = (c^(2))/(a+b)=1`
find `(1)/(1+a) + (1)/(1+b) + (1)/(1 + c)=` ?

Answer

To solve the problem, we start with the given equations:

1. \( a^2 = b + c \)
2. \( b^2 = c + a \)
3. \( c^2 = a + b \)

We can also express these equations in the form:

\[ \frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1 \]

From this, we can derive that:

\[ a^2 = b + c \implies a^2 - b - c = 0 \]
\[ b^2 = c + a \implies b^2 - c - a = 0 \]
\[ c^2 = a + b \implies c^2 - a - b = 0 \]

Next, we need to find the value of:

\[ \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \]

We can start by simplifying each term:

\[ \frac{1}{1+a} = \frac{1}{1+a} \cdot \frac{(1+b)(1+c)}{(1+b)(1+c)} = \frac{(1+b)(1+c)}{(1+a)(1+b)(1+c)} \]

Now, we can express the entire sum:

\[ \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} = \frac{(1+b)(1+c) + (1+a)(1+c) + (1+a)(1+b)}{(1+a)(1+b)(1+c)} \]

Expanding the numerator:

1. \( (1+b)(1+c) = 1 + b + c + bc \)
2. \( (1+a)(1+c) = 1 + a + c + ac \)
3. \( (1+a)(1+b) = 1 + a + b + ab \)

Adding these together:

\[ (1 + b + c + bc) + (1 + a + c + ac) + (1 + a + b + ab) = 3 + 2(a + b + c) + (ab + ac + bc) \]

Now, we can combine everything into our expression:

\[ \frac{3 + 2(a + b + c) + (ab + ac + bc)}{(1+a)(1+b)(1+c)} \]

Next, we need to evaluate \( (1+a)(1+b)(1+c) \):

\[ (1+a)(1+b)(1+c) = 1 + (a+b+c) + (ab + ac + bc) + abc \]

Now we can substitute back into our expression:

\[ \frac{3 + 2(a + b + c) + (ab + ac + bc)}{1 + (a+b+c) + (ab + ac + bc) + abc} \]

To find a specific numerical value, we can use the relationships established by the original equations.

Given that \( a^2 = b + c \), \( b^2 = c + a \), and \( c^2 = a + b \), we can assume \( a = b = c \) for simplicity. Let \( a = b = c = k \). Then we have:

\[ k^2 = k + k \implies k^2 = 2k \implies k(k - 2) = 0 \]

Thus, \( k = 0 \) or \( k = 2 \).

If \( k = 0 \):

\[ \frac{1}{1+0} + \frac{1}{1+0} + \frac{1}{1+0} = 1 + 1 + 1 = 3 \]

If \( k = 2 \):

\[ \frac{1}{1+2} + \frac{1}{1+2} + \frac{1}{1+2} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \]

Thus, the possible values for \( \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \) are either \( 3 \) or \( 1 \).

**Final Answer:**

\[ \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} = 1 \text{ (if } a = b = c = 2 \text{)} \]