Find the product $(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$

Question

Let $1,$ $a_i$ for $1 \leq i \leq 6$ be the different roots of $x^7-1$. Then find the product: $(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$

I don't know how to proceed.

Answer 1

Hint. Factorising, $$x^7-1=(x-1)(x-a_1)\cdots(x-a_6)\ .$$ Dividing by $x-1$ gives $$x^6+x^5+\cdots+1=(x-a_1)\cdots(x-a_6)\ .$$ I'm sure you can finish the problem from here.

Answer 2

$$x^7-1=\prod_{i=1}^7 (x-a_i)$$ with $a_7=1$. So the product is $$\lim_{x\to 1}\frac{x^7-1}{x-1}=7$$

Answer 3

We have $$(x-1)(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)=(x-1)(1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}).$$ Hence we must have $$(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)=1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}.$$ This is because I am dividing the same monomial $x-1$ from both sides. Hence we have $$\prod_{i=1}^{6}(1-a_i)=7.$$