If $x_1,x_2,x_3,\dots ,x_n$ are the roots of the equation $x^n+ax+b=0$, the value of $\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right)\dots \left(x_1-x_n\right)$ is

$$\begin{gathered} \because x^n+ax+b=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)..\left(x-x_n\right) \\ \Rightarrow \left(x-x_2\right)\left(x-x_3\right)\dots \left(x-x_n\right)=\frac{x^n+ax+b}{x-x_1} \end{gathered}$$

On taking $\underset{x\to x_1}{lim}$both sides, we get

$\begin{array}{l}\left(x_1-x_2\right)\left(x_1-x_3\right)\dots \left(x_1-x_n\right)=\underset{x\to x_1}{lim} \frac{x^n+ax+b}{x-x_1}\left[\frac{0}{0}\text{ form }\right]\\ =\underset{x\to x_1}{lim} \frac{n{x}^{n-1}+a}{1}=n{\left(x_1\right)}^{n-1}+a\end{array}$