Without expanding determinant at any stage evaluate `|(1,1,1),(a,b,c),(a^(2)-bc,b^(2)-ca,c^(2)-ab)|`

Answer

To evaluate the determinant \( D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 - bc & b^2 - ca & c^2 - ab \end{vmatrix} \) without expanding it, we will use properties of determinants.

### Step-by-step Solution:

1. **Write the Determinant**:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 - bc & b^2 - ca & c^2 - ab \end{vmatrix} \]

2. **Apply Column Operations**:
We will perform the following operations:
- \( C_2 \leftarrow C_2 - C_1 \)
- \( C_3 \leftarrow C_3 - C_1 \)

This gives us:
\[ D = \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^2 - bc & (b^2 - ca) - (a^2 - bc) & (c^2 - ab) - (a^2 - bc) \end{vmatrix} \]

3. **Simplify the Second and Third Columns**:
The second column becomes:
\[ D = \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^2 - bc & b^2 - a^2 + bc - ca & c^2 - a^2 + bc - ab \end{vmatrix} \]

4. **Factor Out Common Terms**:
Notice that \( b^2 - a^2 = (b - a)(b + a) \) and \( c^2 - a^2 = (c - a)(c + a) \). We can factor these out:
\[ D = (b - a)(c - a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix} \]

5. **Evaluate the Determinant**:
The determinant of a matrix with a row of zeros is zero:
\[ D = (b - a)(c - a) \cdot 0 = 0 \]

### Final Result:
Thus, the value of the determinant is:
\[ \boxed{0} \]