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Question Number 151699 by iloveisrael last updated on 22/Aug/21

$$\sum _{x^3+2022x-2021=0}\left(\frac{1+x}{1-x}\right)=?$$

Answered by mr W last updated on 22/Aug/21

$$x^3+2022x-2021=0$$$${(x-1+1)}^3+2022(x-1+1)-2021=0$$$$letx-1=t$$$${(t+1)}^3+2022(t+1)-2021=0$$$$t^3+3t^2+3t+1+2022t+2022-2021=0$$$$t^3+3t^2+2025t+2=0$$$$\frac{2}{t^3}+\frac{2025}{t^2}+\frac{3}{t}+1=0$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{t}=-\frac{2025}{2}$$$$\frac{1+x}{1-x}=-\frac{x-1+2}{x-1}=-1-\frac{2}{x-1}=-1-\frac{2}{t}$$$$\mathrm{\Sigma }\frac{1+x}{1-x}=\mathrm{\Sigma }(-1-\frac{2}{t})=-3-2\mathrm{\Sigma }\frac{1}{t}$$$$=-3-2\times (-\frac{2025}{2})$$$$=2022$$

Commented by iloveisrael last updated on 22/Aug/21

$$thankyou$$

Answered by aleks041103 last updated on 22/Aug/21

$$\frac{1+x}{1-x}=\frac{1+1-(1-x)}{1-x}=\frac{2}{1-x}-1$$$$x^3+2022x-2021=0has3rootsby$$$$thefundamentaltheoremofalgebra.$$$$\Rightarrow \underset{i=1}{\sum ^3}(\frac{2}{1-x_i}-1)=2\left(\underset{i=1}{\sum ^3}\frac{1}{1-x_i}\right)-3$$$$\underset{i=1}{\sum ^3}\frac{1}{1-x_i}=\frac{(1-x_2)(1-x_3)+(1-x_1)(1-x_3)+(1-x_1)(1-x_2)}{(1-x_1)(1-x_2)(1-x_3)}=$$$$=\frac{3-2(x_1+x_2+x_3)+(x_1{x}_2+x_2{x}_3+x_1{x}_3)}{(1-x_1)(1-x_2)(1-x_3)}$$$$Since{x}_{1,2,3}arerootsofthecubicpolynomial$$$$then$$$$x^3+2022x-2021=(x-x_1)(x-x_2)(x-x_3)$$$$Then$$$$(1-x_1)(1-x_2)(1-x_3)=1^3+2022.1-2021=2$$$$Alsoby{Vieta}^{\prime }sformulas$$$$x_1+x_2+x_3=0$$$$x_1{x}_2+x_2{x}_3+x_1{x}_3=2022$$$$Therefore$$$$\underset{i=1}{\sum ^3}\frac{1}{1-x_i}=\frac{3-2\left(0\right)+2022}{2}=\frac{2025}{2}$$$$Then:$$$$\sum _{x^3+2022x-2021=0}\left(\frac{1+x}{1-x}\right)=2022$$

Commented by iloveisrael last updated on 22/Aug/21

$$thankyou$$

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