If `Delta_1=|(x,a,b),(b,x,a),(a,b,x)| and Delta_2=|(x,b),(a,x)|` are the given determinants then

Options

A. `Delta_1=3(Delta_2)^2`
B. `(d/(dx))(Delta_1)=3Delta_2`
C. `(d/(dx))(Delta_1)=3(Delta_2)_2`
D. `Delta_1=3(Delta_2)^(3/2)`

Answer

To solve the problem, we need to calculate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then analyze the relationship between them.

### Step 1: Calculate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} x & a & b \\ b & x & a \\ a & b & x \end{vmatrix} \]

To calculate this determinant, we can use the method of cofactor expansion along the first row.

\[ \Delta_1 = x \begin{vmatrix} x & a \\ b & x \end{vmatrix} - a \begin{vmatrix} b & a \\ a & x \end{vmatrix} + b \begin{vmatrix} b & x \\ a & b \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} x & a \\ b & x \end{vmatrix} = x^2 - ab \)

2. \( \begin{vmatrix} b & a \\ a & x \end{vmatrix} = bx - a^2 \)

3. \( \begin{vmatrix} b & x \\ a & b \end{vmatrix} = bb - ax = b^2 - ax \)

Substituting these back into the expression for \( \Delta_1 \):

\[ \Delta_1 = x(x^2 - ab) - a(bx - a^2) + b(b^2 - ax) \]

Expanding this:

\[ = x^3 - abx - abx + a^3 + b^3 - abx \]

Combining like terms gives:

\[ \Delta_1 = x^3 + a^3 + b^3 - 3abx \]

### Step 2: Calculate \( \Delta_2 \)

Given:
\[ \Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_2 = x^2 - ab \]

### Step 3: Analyze the relationship between \( \Delta_1 \) and \( \Delta_2 \)

We need to find the derivative of \( \Delta_1 \) with respect to \( x \):

\[ \frac{d}{dx} \Delta_1 = \frac{d}{dx}(x^3 + a^3 + b^3 - 3abx) \]

Calculating the derivative:

\[ = 3x^2 - 3ab \]

Now, we calculate \( 3 \Delta_2 \):

\[ 3 \Delta_2 = 3(x^2 - ab) = 3x^2 - 3ab \]

### Conclusion

From the calculations, we see that:

\[ \frac{d}{dx} \Delta_1 = 3 \Delta_2 \]

Thus, the correct option is that the derivative of \( \Delta_1 \) with respect to \( x \) is equal to 3 times \( \Delta_2 \).