If `|{:(x^(2) +x , 3x - 1 , -x + 3),(2x +1 , 2 + x^(2) , x^(3) - 3),(x - 3, x^(2) + 4, 3x):}| = a_(0) + a_(1) x + a_(2) x^(2) + .... + x_(7) x^(7),` then the value of `a_(0)` is

Options

A. 25
B. 24
C. 23
D. 21

Answer

To solve the given determinant and find the value of \( a_0 \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:

\[ D = \begin{vmatrix} x^2 + x & 3x - 1 & -x + 3 \\ 2x + 1 & 2 + x^2 & x^3 - 3 \\ x - 3 & x^2 + 4 & 3x \end{vmatrix} \]

### Step 2: Substitute \( x = 0 \)
Since we are interested in \( a_0 \), we can substitute \( x = 0 \) into the determinant to simplify our calculations:

\[ D(0) = \begin{vmatrix} 0^2 + 0 & 3(0) - 1 & -0 + 3 \\ 2(0) + 1 & 2 + 0^2 & 0^3 - 3 \\ 0 - 3 & 0^2 + 4 & 3(0) \end{vmatrix} \]

This simplifies to:

\[ D(0) = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 2 & -3 \\ -3 & 4 & 0 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now we will calculate the determinant \( D(0) \) using cofactor expansion along the first row:

\[ D(0) = 0 \cdot \begin{vmatrix} 2 & -3 \\ 4 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 \\ -3 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & -3 \\ -3 & 0 \end{vmatrix} = (1)(0) - (-3)(-3) = 0 - 9 = -9 \)
2. \( \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} = (1)(4) - (2)(-3) = 4 + 6 = 10 \)

Substituting back into the determinant calculation:

\[ D(0) = 0 + 9 + 3 \cdot 10 = 9 + 30 = 39 \]

### Step 4: Conclusion
Thus, the value of \( a_0 \) is:

\[ \boxed{39} \]