Question

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

Answer 1

Answer: 71

Differentiating both sides, we get,
$4p{x}^3+3q{x}^2+2xr+s$ $=\left|\begin{array}{ccc}2x+3& 1& 1\\ x+1& -2x& x-4\\ x-3& x+4& 3x\end{array}\right|+\left|\begin{array}{ccc}{x}^2+3x& x-1& x+3\\ 1& -2& 1\\ x-3& x+4& 3x\end{array}\right|+\left|\begin{array}{ccc}{x}^2+3x& x-1& x+3\\ x+1& -2x& x-4\\ 1& 1& 3\end{array}\right|$
Putting $x=0$ , we get, $s=\left|\begin{array}{ccc}3& 1& 1\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|+\left|\begin{array}{ccc}0& -1& 3\\ 1& -2& 1\\ -3& 4& 0\end{array}\right|+\left|\begin{array}{ccc}0& -1& 3\\ 1& 0& -4\\ 1& 1& 3\end{array}\right|$
$s=64-3+10=71$