If $\alpha_i$ are the roots of $x^n + nax−b = 0$ then show that $\prod_{1< i \le n} (\alpha_1 -\alpha_i)=n(\alpha_1^{n-1}+a)$

Question

If $\alpha_i$ are the roots of $x^n + nax−b = 0$ then I would like to show that $$\prod_{1< i \le n} (\alpha_1 -\alpha_i)=n(\alpha_1^{n-1}+a).$$

The only thing I could think is differentiating $x^n + nax−b = 0.$

What to do after that is the problem.

Answer 1

From the identity $$ x^n + nax + b = \prod_{1\le i \le n} (x -\alpha_i) $$ one has, for $x \neq \alpha_1$, $$ \frac{x^n + nax + b}{x-\alpha_1} = \prod_{1< i \le n} (x -\alpha_i) $$ then letting $x \to \alpha_1$, one gets $$ n\alpha_1^{n-1}+n a=\prod_{1< i \le n} (\alpha_1 -\alpha_i) $$ as announced, where we have used $$ \lim_{x \to \alpha_1}\frac{x^n + nax + b}{x-\alpha_1} =\lim_{x \to \alpha_1}\frac{(x^n + nax + b)-0}{x-\alpha_1} =\lim_{x \to \alpha_1}\frac{f(x)-f(\alpha_1)}{x-\alpha_1}=f'(\alpha_1). $$