Question

If the roots of the equation $x^n-1=0$ are $1,\alpha ,\beta ,\gamma ,...,$ show that $(1-a)(1-\beta )(1-\gamma )....=n$.

Answer 1

Given : Roots of the equation $x^n-1=0$ are $1,\alpha ,\beta ,\gamma ,...$

$\therefore x^n-1=(x-1)(x-\alpha )(x-\beta )\dots$

$\Rightarrow \frac{x^n-1}{x-1}=(x-\alpha )(x-\beta )(x-\gamma )\dots$

Take $x\to 1$

$\Rightarrow \underset{x\to 1}{lim}\frac{x^n-1}{x-1}=n$ .... (using L'hospital's rule)

$\therefore (1-\alpha )(1-\beta )(1-\gamma )=n$