Determinant

\hspace{-20}$Given $\bf{\begin{vmatrix} \bf{a} & \bf{x} & \bf{x} & \bf{x}\\ \bf{x}& \bf{b} & \bf{x} & \bf{x}\\ \bf{x}& \bf{x} & \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{x} & \bf{d} \end{vmatrix} = xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ Now Using property of Determinant, Sub. $\bf{(2)}$ Row from $\bf{(1)}$\\\\\\ $\bf{\Rightarrow \begin{vmatrix} \bf{a-x} & \bf{x-b} & \bf{0} & \bf{0}\\ \bf{x}& \bf{b} & \bf{x} & \bf{x}\\ \bf{x}& \bf{x} & \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{x} & \bf{d} \end{vmatrix}\; = xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ $\bf{\Rightarrow (a-x)\begin{vmatrix} \bf{b} & \bf{x} & \bf{x}\\ \bf{x}& \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{d} \end{vmatrix}-(x-b)\begin{vmatrix} \bf{x} & \bf{x} & \bf{x}\\ \bf{x}& \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{d} \end{vmatrix}=xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ Now after opening the above Determinants......., we get\\\\\\ $\bf{\Rightarrow -3x^4+2x^3(a+b+c+d)-x^2(ab+ac+ad+bc+bd+cd)+abcd=xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\

\hspace{-20}$Now Let we assume $\bf{f(x)=px^4+qx^3+rx^2+sx+t}$\\\\\\ Then $\bf{f\hspace{-10}\quad'(x)=4px^3+3px^2+2rx+s}$\\\\\\ Now put into equation... and Camparing...\\\\\\ $\bf{\Rightarrow p=1\;\;,q=-(a+b+c+d)\;\;,r=(ab+ac+ad+bc+bd+cd)}$\\\\\\ and $\bf{s=0\;\;,t=abcd}$\\\\\\ Using Similarity of $\bf{f(x)-xf\hspace{-10}\quad'(x)}$ to the Determinants, we take $\bf{t=0}$\\\\\\ So $\bf{f(x)=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2+abcd.}$