If `a_1,a_2,a_3 and b_1,b_2,b_3` all `in R` be such that product of any member of first set with any member of the other set is not equal to 1 , then
`Delta=|((1-a_1^3b_1^3)/(1-a_1b_1),(a_1,b_2),(a_1,b_3)),((a_2,b_1),(a_2,b_2),(a_2,b_3)""),((a_3,b_1),(a_3,b_2),(a_3,b_3)"")|` is

Answer

To solve the given determinant problem, we start by rewriting the determinant \(\Delta\) as follows:

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

### Step 1: Simplifying the First Row
The first element in the first row can be simplified using the identity for the difference of cubes:

\[ 1 - a_1^3 b_1^3 = (1 - a_1 b_1)(1 + a_1 b_1 + (a_1 b_1)^2) \]

Thus, we can rewrite the first row as:

\[ \frac{1 - a_1^3 b_1^3}{1 - a_1 b_1} = 1 + a_1 b_1 + (a_1 b_1)^2 \]

So, the determinant becomes:

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

### Step 2: Factor Out Common Terms
Next, we can factor out \(a_2\) from the second row and \(a_3\) from the third row:

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

### Step 3: Recognizing the Structure
Notice that the second and third rows are identical, which means that the determinant evaluates to zero:

\[ \Delta = 0 \]

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

\[ \Delta = 0 \]