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

Commented by Engr_Jidda last updated on 28/Jul/22

$$findthevalueoft$$

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

$$theideaistotransforme$$$$65536t^3+t^2-131076to$$$$theexpressionlike$$$$Z^3-pZ+q=0$$$${256}^2{t}^3+t^2-2({256}^2+2)=0$$$$a={256}^2$$$$t^3+\frac{t^2}{a^2}-2(a+2)=0$$$$t^3+3\left(\frac{1}{3a^2}\right)t^2-2(a+2)=0$$$${(t+\frac{1}{3a^2})}^3-3\left(\frac{1}{9a^4}\right)t-\frac{1}{27a^6}-2(a+2)=0$$$${(t+\frac{1}{3a^2})}^3-\frac{1}{3a^4}(t+\frac{1}{3a^2})+\frac{1}{9a^6}-\frac{1}{27a^6}-2(a+2)=0$$$${(t+\frac{1}{3a^2})}^3-\frac{1}{3a^4}(t+\frac{1}{3a^2})+\frac{2}{27a^6}-2(a+2)=0$$$$Z=(t+\frac{1}{3a^2})=t+\frac{1}{196608}witha=256$$$$Z^3-\frac{1}{3a^2}Z+2(\frac{1}{27a^6}-a-2)=0$$$$p=\frac{1}{3a^2}q=2(\frac{1}{27a^6}-a-2)$$$$theresttocontinue.....$$$$thepricessusislong!$$$$excuseme.$$$$$$$$$$

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

$$\frac{1}{t^3}-\frac{1}{131076t}-\frac{65536}{131076}=0$$$$\frac{1}{t}=\sqrt[3]{\sqrt{{\left(\frac{32768}{131076}\right)}^2-\frac{1}{{393228}^3}}+\frac{32768}{131076}}-\sqrt[3]{\sqrt{{\left(\frac{32768}{131076}\right)}^2-\frac{1}{{393228}^3}}-\frac{32768}{131076}}$$$$\approx 0.793695756$$$$\Rightarrow t\approx 1.259928780$$

Commented by Tawa11 last updated on 31/Jul/22

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

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