Question-174288

Question Number 174288 by Engr_Jidda last updated on 28/Jul/22

Commented by Engr_Jidda last updated on 28/Jul/22

$${find}\:{the}\:{value}\:{of}\:{t} \\ $$

Commented by a.lgnaoui last updated on 30/Jul/22

the idea is to transforme 65536t^3 +t^2 −131076 to the expression like Z^3 −pZ+q=0 256^2 t^3 +t^2 −2(256^2 +2)=0 a=256^2 t^3 +(t^2 /a^2 )−2(a+2)=0 t^3 +3((1/(3a^2 )))t^2 −2(a+2)=0 (t+(1/(3a^2 )))^3 −3((1/(9a^4 )))t−(1/(27a^6 ))−2(a+2)=0 (t+(1/(3a^2 )))^3 −(1/(3a^4 ))(t+(1/(3a^2 )))+(1/(9a^6 ))−(1/(27a^6 ))−2(a+2)=0 (t+(1/(3a^2 )))^3 −(1/(3a^4 ))(t+(1/(3a^2 )))+(2/(27a^6 ))−2(a+2)=0 Z=(t+(1/(3a^2 )))=t+(1/(196608)) with a=256 Z^3 −(1/(3a^2 ))Z+2((1/(27a^6 ))−a−2)=0 p=(1/(3a^2 )) q=2((1/(27a^6 ))−a−2) the rest to continue..... the pricessus is long! excuse me .

$${the}\:{idea}\:{is}\:{to}\:{transforme} \\ $$$$\mathrm{65536}{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −\mathrm{131076}\:\:\:{to} \\ $$$${the}\:{expression}\:{like} \\ $$$${Z}^{\mathrm{3}} −{pZ}+{q}=\mathrm{0} \\ $$$$\mathrm{256}^{\mathrm{2}} {t}^{\mathrm{3}} +{t}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{256}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${a}=\mathrm{256}^{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} +\frac{{t}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{2}\left({a}+\mathrm{2}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right){t}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{9}{a}^{\mathrm{4}} }\right){t}−\frac{\mathrm{1}}{\mathrm{27}{a}^{\mathrm{6}} }−\mathrm{2}\left({a}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{4}} }\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{9}{a}^{\mathrm{6}} }−\frac{\mathrm{1}}{\mathrm{27}{a}^{\mathrm{6}} }−\mathrm{2}\left({a}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{4}} }\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)+\frac{\mathrm{2}}{\mathrm{27}{a}^{\mathrm{6}} }−\mathrm{2}\left({a}+\mathrm{2}\right)=\mathrm{0} \\ $$$${Z}=\left({t}+\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\right)={t}+\frac{\mathrm{1}}{\mathrm{196608}}\:\:\:{with}\:{a}=\mathrm{256} \\ $$$${Z}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }{Z}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{27}{a}^{\mathrm{6}} }−{a}−\mathrm{2}\right)=\mathrm{0} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:{q}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{27}{a}^{\mathrm{6}} }−{a}−\mathrm{2}\right) \\ $$$${the}\:\:{rest}\:{to}\:{continue}….. \\ $$$${the}\:{pricessus}\:{is}\:{long}! \\ $$$${excuse}\:{me}\:. \\ $$$$ \\ $$$$ \\ $$

Answered by mr W last updated on 30/Jul/22

(1/t^3 )−(1/(131076t))−((65536)/(131076))=0 (1/t)=(((√((((32768)/(131076)))^2 −(1/(393228^3 ))))+((32768)/(131076))))^(1/3) −(((√((((32768)/(131076)))^2 −(1/(393228^3 ))))−((32768)/(131076))))^(1/3) ≈0.793 695 756 ⇒t≈1.259 928 780

$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{131076}{t}}−\frac{\mathrm{65536}}{\mathrm{131076}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{t}}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{32768}}{\mathrm{131076}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{393228}^{\mathrm{3}} }}+\frac{\mathrm{32768}}{\mathrm{131076}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{32768}}{\mathrm{131076}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{393228}^{\mathrm{3}} }}−\frac{\mathrm{32768}}{\mathrm{131076}}} \\ $$$$\:\:\:\:\:\approx\mathrm{0}.\mathrm{793}\:\mathrm{695}\:\mathrm{756} \\ $$$$\Rightarrow{t}\approx\mathrm{1}.\mathrm{259}\:\mathrm{928}\:\mathrm{780} \\ $$

Commented by Tawa11 last updated on 31/Jul/22

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

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