If ` (a^(2))/( b + c) = ( b^(2))/( c + a) = ( c^(2))/( a + b)` = 1 then find the value of
`(2)/( 1 + a) + (2)/( 1 + b) + (2)/( 1 + c)`

Options

A. 0
B. 1
C. 2
D. 3

Answer

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

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

This means we can set each fraction equal to 1:

1. From \(\frac{a^2}{b+c} = 1\), we have:
\[ a^2 = b + c \]

2. From \(\frac{b^2}{c+a} = 1\), we have:
\[ b^2 = c + a \]

3. From \(\frac{c^2}{a+b} = 1\), we have:
\[ c^2 = a + b \]

Now we have a system of three equations:

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

Next, we can add all three equations together:

\[ a^2 + b^2 + c^2 = (b+c) + (c+a) + (a+b) \]

This simplifies to:

\[ a^2 + b^2 + c^2 = 2(a + b + c) \]

Rearranging gives us:

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

This can be factored as:

\[ (a-1)^2 + (b-1)^2 + (c-1)^2 = 0 \]

Since the sum of squares is zero, it follows that:

\[ a - 1 = 0, \quad b - 1 = 0, \quad c - 1 = 0 \]

Thus, we conclude that:

\[ a = 1, \quad b = 1, \quad c = 1 \]

Now we can substitute these values into the expression we need to evaluate:

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

Substituting \(a = 1\), \(b = 1\), and \(c = 1\):

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

Thus, the final answer is:

\[ \boxed{3} \]