Question:

Step 1: Determine the constant term $a_7$. The constant term $a_7$ of the polynomial $\Delta(x)$ is obtained by evaluating $\Delta(0)$. $$ \Delta(0) = \begin{vmatrix} 0^2+0 & 2(0)-1 & 0+3 \\ 3(0)+1 & 0^2+2 & 0^3-3 \\ 0-3 & 0^2+4 & 2(0) \end{vmatrix} = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 2 & -3 \\ -3 & 4 & 0 \end{vmatrix} $$ Expanding the determinant along the first row: $$ \Delta(0) = 0(2 \cdot 0 - (-3) \cdot 4) - (-1)(1 \cdot 0 - (-3) \cdot (-3)) + 3(1 \cdot 4 - 2 \cdot (-3)) $$ $$ \Delta(0) = 0 - (-1)(0 - 9) + 3(4 + 6) $$ $$ \Delta(0) = -9 + 30 = 21 $$ Thus, $a_7 = 21$. Step 2: Evaluate $\Delta(1)$. The sum of all coefficients $a_0 + a_1 + \ldots + a_7$ is given by $\Delta(1)$. $$ \Delta(1) = \begin{vmatrix} 1^2+1 & 2(1)-1 & 1+3 \\ 3(1)+1 & 1^2+2 & 1^3-3 \\ 1-3 & 1^2+4 & 2(1) \end{vmatrix} = \begin{vmatrix} 2 & 1 & 4 \\ 4 & 3 & -2 \\ -2 & 5 & 2 \end{vmatrix} $$ Expanding the determinant along the first row: $$ \Delta(1) = 2(3 \cdot 2 - (-2) \cdot 5) - 1(4 \cdot 2 - (-2) \cdot (-2)) + 4(4 \cdot 5 - 3 \cdot (-2)) $$ $$ \Delta(1) = 2(6 + 10) - 1(8 - 4) + 4(20 + 6) $$ $$ \Delta(1) = 2(16) - 1(4) + 4(26) $$ $$ \Delta(1) = 32 - 4 + 104 = 132 $$ Thus, $\sum_{k=0}^{7} a_k = 132$. Step 3: Compute $\sum_{k=0}^{6} a_k$. The sum of coefficients $\sum_{k=0}^{6} a_k$ is given by $\Delta(1) - a_7$. $$ \sum_{k=0}^{6} a_k = \Delta(1) - a_7 = 132 - 21 = 111 $$ Step 4: Evaluate $\Delta(-1)$. $$ \Delta(-1) = \begin{vmatrix} (-1)^2+(-1) & 2(-1)-1 & (-1)+3 \\ 3(-1)+1 & (-1)^2+2 & (-1)^3-3 \\ (-1)-3 & (-1)^2+4 & 2(-1) \end{vmatrix} = \begin{vmatrix} 0 & -3 & 2 \\ -2 & 3 & -4 \\ -4 & 5 & -2 \end{vmatrix} $$ Expanding the determinant along the first row: $$ \Delta(-1) = 0(3 \cdot (-2) - (-4) \cdot 5) - (-3)((-2) \cdot (-2) - (-4) \cdot (-4)) + 2((-2) \cdot 5 - 3 \cdot (-4)) $$ $$ \Delta(-1) = 0 - (-3)(4 - 16) + 2(-10 + 12) $$ $$ \Delta(-1) = 3(-12) + 2(2) $$ $$ \Delta(-1) = -36 + 4 = -32 $$