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Question Number 174304 by dragan91 last updated on 29/Jul/22

Answered by mr W last updated on 29/Jul/22

$$x^5-2x^4-1=0$$$$\mathrm{\Sigma }r=2$$$$\mathrm{\Sigma }{r}_i{r}_j=0$$$${\left(\mathrm{\Sigma }r\right)}^2=\mathrm{\Sigma }{r}^2+2\mathrm{\Sigma }{r}_i{r}_j$$$$2^2=\mathrm{\Sigma }{r}^2+2\times 0$$$$\Rightarrow \mathrm{\Sigma }{r}^2=4$$$$$$$$\frac{1}{x^8}=\frac{1}{x^3}-\frac{2}{x^4}$$$$\frac{1}{x^4}=x-2$$$$\frac{1}{x^3}=x^2-2x$$$$\mathrm{\Sigma }\frac{1}{r^8}=\mathrm{\Sigma }\frac{1}{r^3}-2\mathrm{\Sigma }\frac{1}{r^4}$$$$\mathrm{\Sigma }\frac{1}{r^8}=\mathrm{\Sigma }{r}^2-2\mathrm{\Sigma }r-2(\mathrm{\Sigma }r-10)$$$$\mathrm{\Sigma }\frac{1}{r^8}=\mathrm{\Sigma }{r}^2-4\mathrm{\Sigma }r+20$$$$\mathrm{\Sigma }\frac{1}{r^8}=4-4\times 2+20=16✓$$

Commented by Tawa11 last updated on 29/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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