If `Delta_(1)=|(x,b,b),(a,x,b),(a,a,x)|" and "Delta_(2)=|(x,b),(a,x)|`, then

Options

A. `Delta_(1)=3(Delta_(2))^(2)`
B. `(d)/(dx)(Delta_(1))=3Delta_(2)`
C. `(d)/(dx)(Delta_(1))=3Delta_(2)^(2)`
D. `Delta_(1)=3Delta_(2)^(3//2)`

Explanation

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and find the relationship between them.

### Step-by-Step Solution:

1. **Evaluate \( \Delta_1 \)**:
\[ \Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix} \]
We can expand this determinant using the first row:
\[ \Delta_1 = x \begin{vmatrix} x & b \\ a & x \end{vmatrix} - b \begin{vmatrix} a & b \\ a & x \end{vmatrix} + b \begin{vmatrix} a & x \\ a & a \end{vmatrix} \]

2. **Calculate the 2x2 determinants**:
- First determinant:
\[ \begin{vmatrix} x & b \\ a & x \end{vmatrix} = x^2 - ab \]
- Second determinant:
\[ \begin{vmatrix} a & b \\ a & x \end{vmatrix} = ax - ab \]
- Third determinant:
\[ \begin{vmatrix} a & x \\ a & a \end{vmatrix} = aa - ax = a^2 - ax \]

3. **Substituting back into \( \Delta_1 \)**:
\[ \Delta_1 = x(x^2 - ab) - b(ax - ab) + b(a^2 - ax) \]
Simplifying this:
\[ \Delta_1 = x^3 - abx - abx + b^2 + ba^2 - abx \]
\[ \Delta_1 = x^3 - 3abx + a^2b + b^2 \]

4. **Evaluate \( \Delta_2 \)**:
\[ \Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix} = x^2 - ab \]

5. **Finding the relationship**:
We need to check if \( \Delta_1 \) is related to \( \Delta_2 \). We can differentiate \( \Delta_1 \) with respect to \( x \):
\[ \frac{d\Delta_1}{dx} = 3x^2 - 3ab \]

6. **Calculate \( 3\Delta_2 \)**:
\[ 3\Delta_2 = 3(x^2 - ab) = 3x^2 - 3ab \]

7. **Conclusion**:
Since \( \frac{d\Delta_1}{dx} = 3\Delta_2 \), we conclude that \( \Delta_1 \) is indeed related to \( \Delta_2 \) through differentiation.

### Final Result:
The relationship we found is:
\[ \frac{d\Delta_1}{dx} = 3\Delta_2 \]