If a,b,c be all distinct and
`|(a^3-1,b^3-1,c^3-1),(a,b,c),(a^2,b^2,c^2)|=0` then

Options

A. `sum ab = 0 `
B. `sum a = 0`
C. `abc = 1`
D. `sum a = 1`

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} a^3 - 1 & b^3 - 1 & c^3 - 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

and we know that \(D = 0\).

### Step 1: Factor the first row
The first row can be factored as follows:

\[ a^3 - 1 = (a - 1)(a^2 + a + 1) \]
\[ b^3 - 1 = (b - 1)(b^2 + b + 1) \]
\[ c^3 - 1 = (c - 1)(c^2 + c + 1) \]

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} (a - 1)(a^2 + a + 1) & (b - 1)(b^2 + b + 1) & (c - 1)(c^2 + c + 1) \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

### Step 2: Factor out the common terms
We can factor out \((a - 1)\), \((b - 1)\), and \((c - 1)\) from the first row:

\[ D = (a - 1)(b - 1)(c - 1) \begin{vmatrix} a^2 + a + 1 & b^2 + b + 1 & c^2 + c + 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

### Step 3: Analyze the determinant
Now we have:

\[ D = (a - 1)(b - 1)(c - 1) \cdot D' \]

where

\[ D' = \begin{vmatrix} a^2 + a + 1 & b^2 + b + 1 & c^2 + c + 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

### Step 4: Set the determinant to zero
Since \(D = 0\), we have two cases to consider:

1. \( (a - 1)(b - 1)(c - 1) = 0 \)
2. \( D' = 0 \)

### Step 5: Solve for distinct values
From the first case, we have:

- \( a = 1 \)
- \( b = 1 \)
- \( c = 1 \)

However, \(a\), \(b\), and \(c\) are distinct, so this case does not yield valid solutions.

### Step 6: Analyze the second determinant
Now we need to analyze \(D'\):

\[ D' = \begin{vmatrix} a^2 + a + 1 & b^2 + b + 1 & c^2 + c + 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

To evaluate \(D'\), we can use properties of determinants or perform row operations. The determinant will be zero if the rows are linearly dependent.

### Step 7: Conclusion
The determinant \(D' = 0\) implies that the rows are linearly dependent, which can happen if:

\[ a + b + c = 0 \quad \text{or} \quad ab + bc + ca = 0 \quad \text{or} \quad abc = 1 \]

Thus, we conclude that the condition \(D = 0\) leads to the relationships among \(a\), \(b\), and \(c\) as stated.