If $x_1,x_2,...,x_{n-1},x_n$ be the roots of the equation $$1 + x + x^2 + ... + x^n = 0$$ and $y_1,y_2,...,y_{n},y_{n+1}$ be those of equation $$1 + x + x^2 + ... + x^{n+1} = 0$$ show that $$(1-x_1)(1-x_2)...(1-x_n)=2\left[\frac{1}{1-y_1}+\frac{1}{1-y_2}+...+\frac{1}{1-y_{n+1}}\right]$$
I couldn't get any clue to solve the question. Any suggestion will be very helpful.
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Here is an alternative approach to evaluate the sum $2\,\sum\limits_{k=1}^{n+1}\,\dfrac{1}{1-y_k}$. Let $\omega:=\exp\left(\dfrac{2\pi\text{i}}{n+2}\right)$. Then, set $y_k:=\omega^k$ for $k=1,2,\ldots,n+1$. Therefore, $$\frac1{1-y_k}+\frac1{1-y_{n+2-k}}=\frac{1}{1-\omega^k}+\frac{1}{1-\bar{\omega}^k}=\frac{\left(1-\omega^k\right)+\left(1-\bar{\omega}^k\right)}{1-\omega^k-\bar{\omega}^k+\omega^k\bar{\omega}^k}\,.$$ That is, $$\frac1{1-y_k}+\frac1{1-y_{n+2-k}}=\frac{2-\omega^k-\bar{\omega}^k}{2-\omega^k-\bar{\omega}^k}=1\,.$$ Consequently, $$2\,\sum_{k=1}^{n+1}\,\frac{1}{1-y_k}=\sum_{k=1}^{n+1}\,\left(\frac{1}{1-y_k}+\frac{1}{1-y_{n+2-k}}\right)=\sum_{k=1}^{n+1}\,1=n+1\,.$$
The hint: $$x^{n+1}-1=(x-1)(x-x_1)(x-x_2)...(x-x_n),$$ which gives $$x^n+x^{n-1}+...+x+1=(x-x_1)(x-x_2)...(x-x_n),$$ which for $x=1$ gives $$n+1=(1-x_1)(1-x_2)...(1-x_n).$$ By the similar way you can calculate the second expression.
Indeed, $\frac{1}{y_1-1}$, $\frac{1}{y_2-1}$,...,$\frac{1}{y_{n+1}-1}$ they are roots of the equation: $$\frac{\left(\frac{1}{x}+1\right)^{n+2}-1}{\frac{1}{x}+1-1}=0$$ or $$(1+x)^{n+2}-x^{n+2}=0$$ or $$(n+2)x^{n+1}+\frac{(n+2)(n+1)}{2}x^n+...=0$$ or $$x^{n+1}+\frac{n+1}{2}x^n+...=0,$$ which gives $$\frac{1}{y_1-1}+\frac{1}{y_2-1}+...+\frac{1}{y_{n+1}-1}=-\frac{n+1}{2}$$ or $$\frac{1}{1-y_1}+\frac{1}{1-y_2}+...+\frac{1}{1-y_{n+1}}=\frac{n+1}{2}$$ and we are done!