If a,b,c are distinct real numbers and `|{:(a,a^(2),a^(4)-1),(b,b^(2),b^(4)-1),(c,c^(2),c^(4)-1):}|=0` then
A) a b c = 1
B ) a b c ( a + b + c ) = 1
C ) a b c ( a + b + c ) = a b + b c + c a
D) none of these

Options

A. A)`abc=1`
B. `B) abc(a+b+c)=1`
C. `C) abc(a+b+c)=ab+bc+ca`
D. D) none of these

Answer

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

\[ D = \begin{vmatrix} a & a^2 & a^4 - 1 \\ b & b^2 & b^4 - 1 \\ c & c^2 & c^4 - 1 \end{vmatrix} \]

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

### Step 1: Rewrite the Determinant
We can rewrite the third column \( (a^4 - 1, b^4 - 1, c^4 - 1) \) as \( (a^4 - 1, b^4 - 1, c^4 - 1) = (a^4, b^4, c^4) - (1, 1, 1) \). Thus, we can express the determinant as:

\[ D = \begin{vmatrix} a & a^2 & a^4 \\ b & b^2 & b^4 \\ c & c^2 & c^4 \end{vmatrix} - \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} \]

### Step 2: Apply Determinant Properties
Using the properties of determinants, we can factor out common terms from the rows. The first determinant can be simplified, but we will focus on the second determinant, which is known to be:

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

### Step 3: Evaluate the First Determinant
The first determinant can be evaluated using the formula for the determinant of a Vandermonde matrix:

\[ \begin{vmatrix} a & a^2 & a^4 \\ b & b^2 & b^4 \\ c & c^2 & c^4 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c) \]

### Step 4: Combine Results
Now, substituting back into our expression for \( D \):

\[ D = (a-b)(b-c)(c-a)(a+b+c) - (a-b)(b-c)(c-a) = 0 \]

Factoring out \( (a-b)(b-c)(c-a) \):

\[ (a-b)(b-c)(c-a) \left( (a+b+c) - 1 \right) = 0 \]

### Step 5: Analyze the Factors
Since \( a, b, c \) are distinct real numbers, \( (a-b)(b-c)(c-a) \neq 0 \). Therefore, we must have:

\[ (a+b+c) - 1 = 0 \implies a+b+c = 1 \]

### Conclusion
Thus, we have shown that:

\[ abc(a+b+c) = abc \cdot 1 = abc \]

This leads us to conclude that the correct option is:

**B) \( abc(a+b+c) = 1 \)**