Let `phi_1(x)=x+a_2, phi_2(x)=x^2+b_1x+b_2, x_1=2,x_2=3` and `x_3=5` and `Delta=|(1,1,1),( phi_1(x_1), phi_1(x_2), phi_1(x_3)),(phi_2(x_1),phi_2(x_2),phi_2(x_3))|` . Find the value of `Delta`.

Answer

To find the value of \( \Delta \), we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \phi_1(x_1) & \phi_1(x_2) & \phi_1(x_3) \\ \phi_2(x_1) & \phi_2(x_2) & \phi_2(x_3) \end{vmatrix} \]

where \( \phi_1(x) = x + a_2 \) and \( \phi_2(x) = x^2 + b_1 x + b_2 \), with \( x_1 = 2 \), \( x_2 = 3 \), and \( x_3 = 5 \).

### Step 1: Calculate \( \phi_1(x) \) values
First, we calculate \( \phi_1(x_1) \), \( \phi_1(x_2) \), and \( \phi_1(x_3) \):

- \( \phi_1(x_1) = \phi_1(2) = 2 + a_2 = 2 + a_2 \)
- \( \phi_1(x_2) = \phi_1(3) = 3 + a_2 = 3 + a_2 \)
- \( \phi_1(x_3) = \phi_1(5) = 5 + a_2 = 5 + a_2 \)

### Step 2: Calculate \( \phi_2(x) \) values
Next, we calculate \( \phi_2(x_1) \), \( \phi_2(x_2) \), and \( \phi_2(x_3) \):

- \( \phi_2(x_1) = \phi_2(2) = 2^2 + b_1 \cdot 2 + b_2 = 4 + 2b_1 + b_2 \)
- \( \phi_2(x_2) = \phi_2(3) = 3^2 + b_1 \cdot 3 + b_2 = 9 + 3b_1 + b_2 \)
- \( \phi_2(x_3) = \phi_2(5) = 5^2 + b_1 \cdot 5 + b_2 = 25 + 5b_1 + b_2 \)

### Step 3: Substitute values into the determinant
Now we substitute these values into the determinant:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 + a_2 & 3 + a_2 & 5 + a_2 \\ 4 + 2b_1 + b_2 & 9 + 3b_1 + b_2 & 25 + 5b_1 + b_2 \end{vmatrix} \]

### Step 4: Simplify the determinant
We can simplify this determinant by performing column operations. Let's subtract the first column from the second and the third columns:

\[ \Delta = \begin{vmatrix} 1 & 0 & 0 \\ 2 + a_2 & 1 & 3 \\ 4 + 2b_1 + b_2 & 5 + b_1 & 21 + 3b_1 \end{vmatrix} \]

### Step 5: Calculate the determinant
Now we can expand the determinant along the first row:

\[ \Delta = 1 \cdot \begin{vmatrix} 1 & 3 \\ 5 + b_1 & 21 + 3b_1 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = 1 \cdot (1 \cdot (21 + 3b_1) - 3 \cdot (5 + b_1)) \]

\[ = 21 + 3b_1 - 15 - 3b_1 = 6 \]

Thus, the value of \( \Delta \) is:

\[ \Delta = 6 \]