Value of the `Delta=|{:(,a^(3)-x,a^(4)-x,a^(5)-x),(,a^(5)-x,a^(6)-x,a^(7)-x),(,a^(7)-a,a^(8)-x,a^(9)-x):}|` then the value of `Delta_(1)-Delta_(2)` is (A) 0 (B) `(a^(3)-1)(a^(6)-1)(a^(9)-1)` (C) `(a^(3)+1)(a^(6)+1)(a^(9)+1)` (D) `a^(15)-1`

Options

A. 0
B. `(a^(3)-1)(a^(6)-1)(a^(9)-1)`
C. `(a^(3)+1)(a^(6)+1)(a^(9)+1)`
D. `a^(15)-1`

Answer

To solve the problem, we need to evaluate the determinant given in the question and find the value of \( \Delta_1 - \Delta_2 \).

### Step-by-Step Solution:

1. **Understanding the Determinant**:
We have the determinant:
\[ \Delta = \begin{vmatrix} a^3 - x & a^4 - x & a^5 - x \\ a^5 - x & a^6 - x & a^7 - x \\ a^7 - x & a^8 - x & a^9 - x \end{vmatrix} \]
We need to find \( \Delta_1 - \Delta_2 \).

2. **Applying Properties of Determinants**:
We can express the determinant in terms of two separate determinants:
\[ \Delta = \begin{vmatrix} a^3 & a^4 & a^5 \\ a^5 & a^6 & a^7 \\ a^7 & a^8 & a^9 \end{vmatrix} - \begin{vmatrix} x & x & x \\ x & x & x \\ x & x & x \end{vmatrix} \]

3. **Evaluating the First Determinant**:
The first determinant can be simplified. Notice that if we multiply the second column by \( a \) and the third column by \( a^2 \), we can factor out \( a \) and \( a^2 \) respectively:
\[ \Delta_1 = a \cdot a^2 \cdot \begin{vmatrix} a^3 & a^4 & a^5 \\ a^5 & a^6 & a^7 \\ a^7 & a^8 & a^9 \end{vmatrix} \]

4. **Recognizing Redundant Columns**:
After the transformation, we observe that the second and third columns become identical, leading to:
\[ \Delta_1 = 0 \]

5. **Evaluating the Second Determinant**:
The second determinant is:
\[ \Delta_2 = \begin{vmatrix} x & x & x \\ x & x & x \\ x & x & x \end{vmatrix} \]
Since all rows are identical, this determinant also evaluates to:
\[ \Delta_2 = 0 \]

6. **Final Calculation**:
Now, we compute \( \Delta_1 - \Delta_2 \):
\[ \Delta_1 - \Delta_2 = 0 - 0 = 0 \]

### Conclusion:
Thus, the value of \( \Delta_1 - \Delta_2 \) is:
\[ \boxed{0} \]