If `x= b+c, y=c+a, z=a+b`, then find the value of `(x^2 + y^2 + z^2 -yz - zx-xy)/( a^(2) + b^(2) + c^(2) - bc - ca - ab)`

Answer

To solve the problem, we need to find the value of

\[ \frac{x^2 + y^2 + z^2 - yz - zx - xy}{a^2 + b^2 + c^2 - bc - ca - ab} \]

given that \(x = b + c\), \(y = c + a\), and \(z = a + b\).

### Step 1: Simplify the Numerator

The numerator is

\[ x^2 + y^2 + z^2 - yz - zx - xy \]

We can use the algebraic identity:

\[ m^2 + n^2 + p^2 - mn - np - pm = \frac{1}{2} \left( (m-n)^2 + (n-p)^2 + (p-m)^2 \right) \]

In our case, let \(m = x\), \(n = y\), and \(p = z\). Thus, we can rewrite the numerator as:

\[ \frac{1}{2} \left( (x-y)^2 + (y-z)^2 + (z-x)^2 \right) \]

### Step 2: Calculate \(x - y\), \(y - z\), and \(z - x\)

Now we need to find \(x - y\), \(y - z\), and \(z - x\):

1. **Calculate \(x - y\)**:
\[ x - y = (b + c) - (c + a) = b - a \]

2. **Calculate \(y - z\)**:
\[ y - z = (c + a) - (a + b) = c - b \]

3. **Calculate \(z - x\)**:
\[ z - x = (a + b) - (b + c) = a - c \]

### Step 3: Substitute Back into the Numerator

Now substituting these values back into the numerator:

\[ \frac{1}{2} \left( (b-a)^2 + (c-b)^2 + (a-c)^2 \right) \]

### Step 4: Simplify the Denominator

The denominator is

\[ a^2 + b^2 + c^2 - bc - ca - ab \]

This expression can also be recognized as:

\[ \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) \]

### Step 5: Putting It All Together

Now we can write the entire expression as:

\[ \frac{\frac{1}{2} \left( (b-a)^2 + (c-b)^2 + (a-c)^2 \right)}{\frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right)} \]

The \(\frac{1}{2}\) cancels out:

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

### Step 6: Conclusion

Since the numerator and denominator are the same, the value of the expression is:

\[ 1 \]