If `Delta=|(x+1,x^2+2,x(x+1)),(x^2+1,x+1,x^2+2),(x^2+2,x(x+1),x+1)| = p_0x^(6) +p_1x^(5) +p_2x^(4)+p_3x^3+p_4x^2+p_5x+p_6` then `(p_5,p_6)` =

Options

A. (-3,-7)
B. (-5,9)
C. (-3,-5)
D. (-3,7)

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} x+1 & x^2+2 & x(x+1) \\ x^2+1 & x+1 & x^2+2 \\ x^2+2 & x(x+1) & x+1 \end{vmatrix} \]

We want to express this determinant as a polynomial of the form:

\[ \Delta = p_0 x^6 + p_1 x^5 + p_2 x^4 + p_3 x^3 + p_4 x^2 + p_5 x + p_6 \]

and find the coefficients \(p_5\) and \(p_6\).

### Step 1: Calculate \(p_6\)

To find \(p_6\), we substitute \(x = 0\) into the determinant:

\[ \Delta(0) = \begin{vmatrix} 0+1 & 0^2+2 & 0(0+1) \\ 0^2+1 & 0+1 & 0^2+2 \\ 0^2+2 & 0(0+1) & 0+1 \end{vmatrix} \]

This simplifies to:

\[ \Delta(0) = \begin{vmatrix} 1 & 2 & 0 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{vmatrix} \]

Now we can calculate this determinant using the formula for a \(3 \times 3\) determinant:

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

Calculating the smaller determinants:

\[ \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 0 = 1 \]

\[ \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = -3 \]

Putting it all together:

\[ \Delta(0) = 1 \cdot 1 - 2 \cdot (-3) = 1 + 6 = 7 \]

Thus, \(p_6 = 7\).

### Step 2: Calculate \(p_5\)

To find \(p_5\), we differentiate the determinant with respect to \(x\) and then substitute \(x = 0\):

\[ \Delta' = \frac{d}{dx} \begin{vmatrix} x+1 & x^2+2 & x(x+1) \\ x^2+1 & x+1 & x^2+2 \\ x^2+2 & x(x+1) & x+1 \end{vmatrix} \]

Using the property of determinants, we differentiate the first row:

\[ \Delta' = \begin{vmatrix} 1 & 2x & (x+1) + x \\ x^2+1 & x+1 & x^2+2 \\ x^2+2 & x(x+1) & x+1 \end{vmatrix} + \begin{vmatrix} x+1 & 2 & x(x+1) \\ x^2+1 & 1 & x^2+2 \\ x^2+2 & x(x+1) & x+1 \end{vmatrix} + \begin{vmatrix} x+1 & x^2+2 & (x+1) \\ x^2+1 & x+1 & 2 \\ x^2+2 & (x+1) & 0 \end{vmatrix} \]

Now we substitute \(x = 0\) into the differentiated determinant:

Calculating the first determinant at \(x = 0\):

\[ \Delta'(0) = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = 1 \cdot \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} - 0 + 0 = 1 \cdot 1 = 1 \]

Calculating the second determinant at \(x = 0\):

\[ = \begin{vmatrix} 1 & 2 & 0 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{vmatrix} = 7 \text{ (as calculated previously)} \]

Calculating the third determinant at \(x = 0\):

\[ = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 2 & 0 & 0 \end{vmatrix} = 0 \]

Thus, we have:

\[ \Delta'(0) = 1 + 7 + 0 = 8 \]

So, \(p_5 = 8\).

### Final Result

The values of \(p_5\) and \(p_6\) are:

\[ (p_5, p_6) = (8, 7) \]