If a,b,c are distinct real numbers and `|{:(a,a^(2),a^(3)-1),(b,b^(2),b^(3)-1),(c,c^(2),c^(3)-1):}|=0` then

Options

A. `a+b+c=0`
B. `abc=1`
C. `a+b+c=1`
D. `ab+bc+ca=0`

Answer

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

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

and show that it equals zero under the condition that \(a\), \(b\), and \(c\) are distinct real numbers.

### Step 1: Write the Determinant
We start with the determinant as given:

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

### Step 2: Expand the Determinant
We can expand the determinant using the properties of determinants. We can separate the last column:

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

### Step 3: Evaluate the First Determinant
The first determinant can be recognized as the Vandermonde determinant:

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

### Step 4: Evaluate the Second Determinant
The second determinant is also a Vandermonde determinant:

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

### Step 5: Combine the Results
Combining the results from the two determinants, we have:

\[ D = (b-a)(c-a)(c-b) - (b-a)(c-a)(c-b) \]

This simplifies to:

\[ D = 0 \]

### Conclusion
Since the determinant \(D\) is equal to zero, we conclude that:

\[ |{(a, a^2, a^3 - 1), (b, b^2, b^3 - 1), (c, c^2, c^3 - 1)}| = 0 \]

This implies that the product of the factors must equal zero, leading us to the conclusion that \(abc = 1\).

### Final Result
Thus, we find that:

\[ abc = 1 \]