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

Commented by Tawa1 last updated on 20/Oct/18

Sir, can you share me the link to download this pdf or how can i get it sir

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

Commented by Kunal12588 last updated on 20/Oct/18

needs imaginary numbers real not working

$${needs}\:{imaginary}\:{numbers}\: \\ $$$${real}\:{not}\:{working} \\ $$

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

i have checked Hall and knight higher algebra and Bernard and child ...found some theorem i have solved the problem tsking help...

$${i}\:{have}\:{checked}\:{Hall}\:{and}\:{knight}\:{higher}\:{algebra} \\ $$$${and}\:{Bernard}\:{and}\:{child}\:...{found}\:{some}\:{theorem} \\ $$$${i}\:{have}\:{solved}\:{the}\:{problem}\:{tsking}\:{help}... \\ $$

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

God bless you sir

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

Answered by MJS last updated on 20/Oct/18

hope this helps: we can write the solutions of any polynome of 3^(rd) degree with real factors as x_1 =t ⇒ x_1 ^5 =t^5 x_2 =re^(iα) ⇒ x_2 ^5 =r^5 e^(5iα) x_3 =re^(−iα) ⇒ x_3 ^5 =r^5 e^(−5iα) (1/x_1 ^5 )+(1/x_2 ^5 )+(1/x_3 ^5 )=t^(−5) +r^(−5) (e^(5iα) +e^(−5iα) )= =(1/t^5 )+(2/r^5 )cos 5α and this is a real number

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

Answered by ajfour last updated on 20/Oct/18

COMMON SENSE Method _(−) x^3 −2x^2 +x+1=0 (1/x^5 ) = (2/x^3 )−(1/x^4 )−(1/x^2 ) Σ(1/p^5 )= 2Σ(1/p^3 )−Σ(1/p^4 )−Σ(1/p^2 ) (1/x^4 )= −(1/x^3 )+(2/x^2 )−(1/x) Σ(1/p^5 )=2Σ(1/p^3 )+(Σ(1/p^3 )−2Σ(1/p^2 )+Σ(1/p)) −Σ(1/p^2 ) Σ(1/p^5 ) = 3Σ(1/p^3 )−3Σ(1/p^2 )+Σ(1/p) (1/x^3 ) = −(1/x^2 )+(2/x)−1 Σ(1/p^5 ) = 3(−Σ(1/p^2 )+2Σ(1/p)−Σ1) −3Σ(1/p^2 )+Σ(1/p) Σ(1/p^5 ) = −6Σ(1/p^2 )+7Σ(1/p)−9 (1/x^2 ) = −(1/x)+2−x Σ(1/p^5 ) = −6(−Σ(1/p)+Σ2−Σp) +7Σ(1/p)−9 = 13Σ(1/p)+6Σp−45 Σ(1/p^5 ) = 13(((pq+qr+rp)/(pqr)))+6(p+q+r)−45 and since pq+qr+rp = 1 p+q+r = 2 ; pqr = −1 , So Σ(1/p^5 ) = 13(−1)+6(2)−45 = −46 .

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

Commented by naka3546 last updated on 20/Oct/18

−46

$$−\mathrm{46} \\ $$

Commented by ajfour last updated on 20/Oct/18

yes, i corrected!

$${yes},\:{i}\:{corrected}! \\ $$

Answered by naka3546 last updated on 20/Oct/18

Commented by Tawa1 last updated on 20/Oct/18

The image is not clear sir

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

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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