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Question Number 46042 by naka3546 last updated on 20/Oct/18

Commented by Tawa1 last updated on 20/Oct/18

$$\mathrm{Sir},\mathrm{can}\mathrm{you}\mathrm{share}\mathrm{me}\mathrm{the}\mathrm{link}\mathrm{to}\mathrm{download}\mathrm{this}\mathrm{pdf}\mathrm{or}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{get}\mathrm{it}\mathrm{sir}$$

Commented by Kunal12588 last updated on 20/Oct/18

$$needsimaginarynumbers$$$$realnotworking$$

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

$$ihavecheckedHallandknighthigheralgebra$$$$andBernardandchild...foundsometheorem$$$$ihavesolvedtheproblemtskinghelp...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by Tawa1 last updated on 20/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 20/Oct/18

$$\mathrm{hope}\mathrm{this}\mathrm{helps}:$$$$\mathrm{we}\mathrm{can}\mathrm{write}\mathrm{the}\mathrm{solutions}\mathrm{of}\mathrm{any}\mathrm{polynome}$$$$\mathrm{of}{3}^{\mathrm{rd}}\mathrm{degree}\mathrm{with}\mathrm{real}\mathrm{factors}\mathrm{as}$$$$x_1=t\Rightarrow x_1^5=t^5$$$$x_2=r{\mathrm{e}}^{\mathrm{i}\alpha }\Rightarrow x_2^5=r^5{\mathrm{e}}^{5\mathrm{i}\alpha }$$$$x_3=r{\mathrm{e}}^{-\mathrm{i}\alpha }\Rightarrow x_3^5=r^5{\mathrm{e}}^{-5\mathrm{i}\alpha }$$$$$$$$\frac{1}{x_1^5}+\frac{1}{x_2^5}+\frac{1}{x_3^5}=t^{-5}+r^{-5}({\mathrm{e}}^{5\mathrm{i}\alpha }+{\mathrm{e}}^{-5\mathrm{i}\alpha })=$$$$=\frac{1}{t^5}+\frac{2}{r^5}\cos 5\alpha \mathrm{and}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}$$

Answered by ajfour last updated on 20/Oct/18

$$\underset{-}{\mathcal{COMMON}\mathcal{SENSE}Method}$$$${\mathit{x}}^3-2{\mathit{x}}^2+\mathit{x}+1=0$$$$\frac{1}{{\mathit{x}}^5}=\frac{2}{{\mathit{x}}^3}-\frac{1}{{\mathit{x}}^4}-\frac{1}{{\mathit{x}}^2}$$$$\mathrm{\Sigma }\frac{1}{{\mathit{p}}^5}=2\mathrm{\Sigma }\frac{1}{p^3}-\mathrm{\Sigma }\frac{1}{p^4}-\mathrm{\Sigma }\frac{1}{p^2}$$$$\frac{1}{x^4}=-\frac{1}{x^3}+\frac{2}{x^2}-\frac{1}{x}$$$$\mathrm{\Sigma }\frac{1}{p^5}=2\mathrm{\Sigma }\frac{1}{p^3}+(\mathrm{\Sigma }\frac{1}{p^3}-2\mathrm{\Sigma }\frac{1}{p^2}+\mathrm{\Sigma }\frac{1}{p})$$$$-\mathrm{\Sigma }\frac{1}{p^2}$$$$\mathrm{\Sigma }\frac{1}{p^5}=3\mathrm{\Sigma }\frac{1}{p^3}-3\mathrm{\Sigma }\frac{1}{p^2}+\mathrm{\Sigma }\frac{1}{p}$$$$\frac{1}{x^3}=-\frac{1}{x^2}+\frac{2}{x}-1$$$$\mathrm{\Sigma }\frac{1}{p^5}=3(-\mathrm{\Sigma }\frac{1}{p^2}+2\mathrm{\Sigma }\frac{1}{p}-\mathrm{\Sigma }1)$$$$-3\mathrm{\Sigma }\frac{1}{p^2}+\mathrm{\Sigma }\frac{1}{p}$$$$\mathrm{\Sigma }\frac{1}{p^5}=-6\mathrm{\Sigma }\frac{1}{p^2}+7\mathrm{\Sigma }\frac{1}{p}-9$$$$\frac{1}{x^2}=-\frac{1}{x}+2-x$$$$\mathrm{\Sigma }\frac{1}{p^5}=-6(-\mathrm{\Sigma }\frac{1}{p}+\mathrm{\Sigma }2-\mathrm{\Sigma }p)$$$$+7\mathrm{\Sigma }\frac{1}{p}-9$$$$=13\mathrm{\Sigma }\frac{1}{p}+6\mathrm{\Sigma }p-45$$$$\mathrm{\Sigma }\frac{1}{p^5}=13\left(\frac{pq+qr+rp}{pqr}\right)+6(p+q+r)-45$$$$andsincepq+qr+rp=1$$$$p+q+r=2;pqr=-1,So$$$$\mathrm{\Sigma }\frac{1}{{\mathit{p}}^5}=13(-1)+6\left(2\right)-45$$$$=-46.$$$$$$

Commented by naka3546 last updated on 20/Oct/18

$$-46$$

Commented by ajfour last updated on 20/Oct/18

$$yes,icorrected!$$

Answered by naka3546 last updated on 20/Oct/18

Commented by Tawa1 last updated on 20/Oct/18

$$\mathrm{The}\mathrm{image}\mathrm{is}\mathrm{not}\mathrm{clear}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

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