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 find the value of the determinant

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

we can simplify the elements of the determinant using the identity for the difference of cubes.

### Step 1: Simplify each element of the determinant
Each element in the determinant can be rewritten using the formula for the difference of cubes:

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

Thus, we can rewrite each element as follows:

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

So, the determinant simplifies to:

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

### Step 2: Identify common patterns
Notice that each column can be expressed in a similar form. This suggests that the columns may be linearly dependent.

### Step 3: Check for linear dependence
To check for linear dependence, we can observe that if we subtract the first column from the second and third columns, we will see that the resulting columns will have a similar structure.

If we perform the operation \( C_2 - C_1 \) and \( C_3 - C_1 \), we will find that the new columns are linear combinations of the original columns.

### Step 4: Conclude the determinant value
Since the determinant of a matrix with linearly dependent columns is zero, we conclude that:

\[ \Delta = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]