The value of the determinant
`Delta = |((1 - a_(1)^(3) b_(1)^(3))/(1 - a_(1) b_(1)),(1 - a_(1)^(3) b_(2)^(3))/(1 - a_(1) b_(2)),(1 - a_(1)^(3) b_(3)^(3))/(1 - a_(1) b_(3))),((1 - a_(2)^(3) b_(1)^(3))/(1 - a_(2) b_(1)),(1 - a_(2)^(3) b_(2)^(3))/(1 - a_(2) b_(2)),(1 - a_(2)^(3) b_(3)^(3))/(1 - a_(2) b_(3))),((1 - a_(3)^(3) b_(1)^(3))/(1 - a_(3) b_(1)),(1 - a_(3)^(3) b_(2)^(3))/(1 - a_(3) b_(2)),(1 - a_(3)^(3) b_(3)^(3))/(1 - a_(3) b_(3)))|`, is

Options

A. 0
B. dependent only on `a_(1), a_(2), a_(3)`
C. dependent only `b_(1), b_(2), b_(3)`
D. dependent on `a_(1), a_(2), a_(3) b_(1), b_(2), b_(3)`

Answer

To solve the determinant given in the question, we will follow these steps:

### Step 1: Rewrite the Determinant
The determinant is given as:

\[ \Delta = \begin{vmatrix} \frac{1 - a_1^3 b_1^3}{1 - a_1 b_1} & \frac{1 - a_1^3 b_2^3}{1 - a_1 b_2} & \frac{1 - a_1^3 b_3^3}{1 - a_1 b_3} \\ \frac{1 - a_2^3 b_1^3}{1 - a_2 b_1} & \frac{1 - a_2^3 b_2^3}{1 - a_2 b_2} & \frac{1 - a_2^3 b_3^3}{1 - a_2 b_3} \\ \frac{1 - a_3^3 b_1^3}{1 - a_3 b_1} & \frac{1 - a_3^3 b_2^3}{1 - a_3 b_2} & \frac{1 - a_3^3 b_3^3}{1 - a_3 b_3} \end{vmatrix} \]

### Step 2: Factor the Numerator
Each term in the determinant can be factored as follows:

\[ 1 - a_i^3 b_j^3 = (1 - a_i b_j)(1 + a_i b_j + a_i^2 b_j^2) \]

Thus, we can rewrite each entry in the determinant:

\[ \Delta = \begin{vmatrix} (1 - a_1 b_1)(1 + a_1 b_1 + a_1^2 b_1^2) & (1 - a_1 b_2)(1 + a_1 b_2 + a_1^2 b_2^2) & (1 - a_1 b_3)(1 + a_1 b_3 + a_1^2 b_3^2) \\ (1 - a_2 b_1)(1 + a_2 b_1 + a_2^2 b_1^2) & (1 - a_2 b_2)(1 + a_2 b_2 + a_2^2 b_2^2) & (1 - a_2 b_3)(1 + a_2 b_3 + a_2^2 b_3^2) \\ (1 - a_3 b_1)(1 + a_3 b_1 + a_3^2 b_1^2) & (1 - a_3 b_2)(1 + a_3 b_2 + a_3^2 b_2^2) & (1 - a_3 b_3)(1 + a_3 b_3 + a_3^2 b_3^2) \end{vmatrix} \]

### Step 3: Simplify the Determinant
Now, we can separate the determinant into two parts:

\[ \Delta = \begin{vmatrix} 1 - a_1 b_1 & 1 - a_1 b_2 & 1 - a_1 b_3 \\ 1 - a_2 b_1 & 1 - a_2 b_2 & 1 - a_2 b_3 \\ 1 - a_3 b_1 & 1 - a_3 b_2 & 1 - a_3 b_3 \end{vmatrix} \cdot \begin{vmatrix} 1 + a_1 b_1 + a_1^2 b_1^2 & 0 & 0 \\ 0 & 1 + a_2 b_2 + a_2^2 b_2^2 & 0 \\ 0 & 0 & 1 + a_3 b_3 + a_3^2 b_3^2 \end{vmatrix} \]

### Step 4: Evaluate the Determinants
The first determinant evaluates to zero if any two rows are linearly dependent, which they are if \( a_i = a_j \) or \( b_i = b_j \) for any \( i \neq j \).

The second determinant is a diagonal matrix and does not affect the zero result from the first determinant.

### Final Result
Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = 0 \]