The value of the determinant `|(1,a,a^2-bc),(1,b,b^2-ca),(1,c,c^2-ab)|` is (A) `(a+b+c),(a^2+b^2+c^2)` (B) `a^3+b^3+c^3-3abc` (C) `(a-b)(b-c)(c-a)` (D) 0

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 1 & a & a^2 - bc \\ 1 & b & b^2 - ca \\ 1 & c & c^2 - ab \end{vmatrix} \]

we will perform some row operations to simplify the determinant.

### Step 1: Subtract the first row from the second and third rows

We will apply the row operations \( R_2 \leftarrow R_2 - R_1 \) and \( R_3 \leftarrow R_3 - R_1 \):

\[ \Delta = \begin{vmatrix} 1 & a & a^2 - bc \\ 0 & b - a & b^2 - ca - (a^2 - bc) \\ 0 & c - a & c^2 - ab - (a^2 - bc) \end{vmatrix} \]

Calculating the entries in the second and third rows:

- For \( R_2 \):
\[ b^2 - ca - (a^2 - bc) = b^2 - ca - a^2 + bc = b^2 - a^2 + bc - ca \]

- For \( R_3 \):
\[ c^2 - ab - (a^2 - bc) = c^2 - ab - a^2 + bc = c^2 - a^2 + bc - ab \]

Thus, we have:

\[ \Delta = \begin{vmatrix} 1 & a & a^2 - bc \\ 0 & b - a & b^2 - a^2 + bc - ca \\ 0 & c - a & c^2 - a^2 + bc - ab \end{vmatrix} \]

### Step 2: Factor out common terms

Now we can factor out \( (b - a) \) and \( (c - a) \) from the second and third rows respectively:

\[ \Delta = (b - a)(c - a) \begin{vmatrix} 1 & a & a^2 - bc \\ 0 & 1 & \frac{b^2 - a^2 + bc - ca}{b - a} \\ 0 & 1 & \frac{c^2 - a^2 + bc - ab}{c - a} \end{vmatrix} \]

### Step 3: Simplify the determinant

Now we can simplify the determinant further. The new determinant is:

\[ \Delta = (b - a)(c - a) \begin{vmatrix} 1 & a & a^2 - bc \\ 0 & 1 & \frac{(b - a)(b + a) + bc - ca}{b - a} \\ 0 & 1 & \frac{(c - a)(c + a) + bc - ab}{c - a} \end{vmatrix} \]

### Step 4: Final determinant evaluation

Since we have two identical rows (the second and third rows are now multiples of each other), the determinant evaluates to zero:

\[ \Delta = 0 \]

### Conclusion

Thus, the value of the determinant is:

\[ \boxed{0} \]