Question

Let $px^{4}+qx^{3}+rx^{2}+sx+t=$ $\begin{vmatrix} x^{2}+3x & x-1 & x+3 \\ x+1 & -2x & x-4 \\ x-3 & x+4 & 3x \end{vmatrix}$ be an identity, where $p,q,r,s$ and $t$ are constants, then the value of $s$ is equal to

Answer 1

Differentiating both sides, we get,
$4px^{3}+3qx^{2}+2xr+s$ $=\begin{vmatrix} 2x+3 & 1 & 1 \\ x+1 & -2x & x-4 \\ x-3 & x+4 & 3x \end{vmatrix}+\begin{vmatrix} x^{2}+3x & x-1 & x+3 \\ 1 & -2 & 1 \\ x-3 & x+4 & 3x \end{vmatrix}+\begin{vmatrix} x^{2}+3x & x-1 & x+3 \\ x+1 & -2x & x-4 \\ 1 & 1 & 3 \end{vmatrix}$
Putting $x=0$ , we get, $s=\begin{vmatrix} 3 & 1 & 1 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix}+\begin{vmatrix} 0 & -1 & 3 \\ 1 & -2 & 1 \\ -3 & 4 & 0 \end{vmatrix}+\begin{vmatrix} 0 & -1 & 3 \\ 1 & 0 & -4 \\ 1 & 1 & 3 \end{vmatrix}$
$s=64-3+10=71$