Question

If $f\left(x\right)=f(x-a)(x-b)(x-c)(x-d)$ then prove that
$\Delta =\left|\begin{array}{cccc}a& x& x& x\\ x& b& x& x\\ x& x& c& x\\ x& x& x& d\end{array}\right|=f\left(x\right)-x{f}^{{}^{\prime }}\left(x\right).$

Answer 1

$f\left(x\right)=(x-a)(x-b)(x-c)(c-d)$
$\log f\left(x\right)=\sum \log (x-a)$
$\frac{1}{f\left(x\right)}.f^{{}^{\prime }}\left(x\right)=\frac{1}{x-a}+..$
Our answer is
$f\left(x\right)-x{f}^{{}^{\prime }}\left(x\right)=f\left(x\right)[1-x.\frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}]$
$f\left(x\right)[1-x\{\frac{1}{x-a}+...\}]$
$\left(\right)\left(\right)\left(\right)\left(\right)-x\left\{\right(\left)\right(\left)\right()+\cdots +\cdots +\}$ .(1)
Apply $C_1-C_2$, $C_2-C_3$, $C_3-C_4$
$\Delta =\left|\begin{array}{cccc}-(x-a)& 0& 0& x\\ (x-b)& -(x-b)& 0& x\\ 0& (x-c)& -(x-c)& x\\ 0& 0& (x-d)& d\end{array}\right|$
Expand with first row
$\Delta =(x-a)\left|\begin{array}{ccc}-(x-b)& 0& x\\ (x-c)& -(x-c)& x\\ 0& (x+d)& d\end{array}\right|-\left|\begin{array}{ccc}(x-b)& -(x-b)& 0\\ 0& (x-c)& -(x-c)\\ 0& 0& (x-d)\end{array}\right|$
$=(x-a)(x-b)\left(\right)+(x-c)d+(x-b)(x-d)x+x(x-c)(x-d)-x(x-b)(x-c)(x-d)$
Now write $d$ as $x-(x-d)$.
It takes the form of $\left(1\right)$ i.e.
$(x-a)(x-b)(x-c)(x-d)-x\{(x-a)(x-b)(x-c)+\cdots +\cdots +\}$
i.e. $\Delta =f\left(x\right)-x{f}^{{}^{\prime }}\left(x\right)$.