If `|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|= (a -b) (b -c) (c -a) (a + b+c)`
where a, b, c are all different, then the determinant
`|(1,1,1),((x-a)^(2),(x -b)^(2),(x -c)^(2)),((x -b) (x -c),(x -c) (x -a),(x -a) (x -b))|` vanishes when

Options

A. `a+b+c=0`
B. `x=(1)/(3) (a+b+c)`
C. `x=(1)/(2) (a+b+c)`
D. `x=a+b+c`

Explanation

To solve the given problem step by step, we need to evaluate the determinant and find the value of \( x \) for which it vanishes.

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ (x-a)^2 & (x-b)^2 & (x-c)^2 \\ (x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b) \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can simplify the determinant by performing column operations. We will multiply the first column by \( (x-a) \), the second column by \( (x-b) \), and the third column by \( (x-c) \). This gives:
\[ D = \frac{1}{(x-a)(x-b)(x-c)} \begin{vmatrix} 1 & 1 & 1 \\ (x-a)^3 & (x-b)^3 & (x-c)^3 \\ (x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b) \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Next, we take out common factors from the rows:
\[ D = \frac{(x-a)(x-b)(x-c)}{(x-a)(x-b)(x-c)} \begin{vmatrix} 1 & 1 & 1 \\ (x-a)^3 & (x-b)^3 & (x-c)^3 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 4: Evaluate the Simplified Determinant
Now, we can evaluate the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ (x-a)^3 & (x-b)^3 & (x-c)^3 \\ 1 & 1 & 1 \end{vmatrix} \]
This determinant can be simplified further by subtracting the first row from the last row:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ (x-a)^3 & (x-b)^3 & (x-c)^3 \\ 0 & 0 & 0 \end{vmatrix} \]
Since the last row is all zeros, the determinant \( D \) vanishes.

### Step 5: Set the Condition for Vanishing
For the determinant to vanish, we set:
\[ D = 0 \]
This implies that the expression \( (x-a)(x-b)(x-c) \) must equal zero, which occurs when \( x = a \), \( x = b \), or \( x = c \).

### Conclusion
Thus, the determinant vanishes when:
\[ x = a, \quad x = b, \quad \text{or} \quad x = c \]