If `(a)/(b+c) + (b)/(c+a) + (c )/(a+b)= 1` find `(a^(2))/(b+c) + (b^(2))/(c+a) + (c^(2))/(a+b)=` ?

Answer

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

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

We need to find the value of:

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

### Step 1: Multiply the entire equation by \( (b+c)(c+a)(a+b) \)

This will eliminate the denominators. We have:

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

### Step 2: Expand both sides

Expanding the left side:

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

Combining these gives us a polynomial in \( a, b, c \).

### Step 3: Simplify the left side

The left side can be rearranged and simplified to yield:

\[ a^2(b+c) + b^2(c+a) + c^2(a+b) + 3abc \]

### Step 4: Expand the right side

The right side expands to:

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

### Step 5: Set the two sides equal

Setting the simplified left side equal to the expanded right side gives us:

\[ a^2(b+c) + b^2(c+a) + c^2(a+b) + 3abc = abc + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc \]

### Step 6: Rearranging terms

Rearranging the equation results in:

\[ a^2(b+c) + b^2(c+a) + c^2(a+b) = abc + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b - 3abc \]

### Step 7: Recognizing the pattern

Notice that the left side is equal to the right side minus \( 3abc \). Since we know from the original equation that the left side equals 1, we can conclude:

\[ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0 \]

### Final Result

Thus, the value of \( \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \) is:

\[ \boxed{0} \]