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Question Number 172326 by nurtani last updated on 25/Jun/22

$$If$$$$(x^3+x^2+x)+(\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x})=70$$$$x^4+\frac{1}{x^4}=?$$

Commented by infinityaction last updated on 25/Jun/22

$$x^3+\frac{1}{x^3}+x^2+\frac{1}{x^2}+x+\frac{1}{x}=70$$$${(x+\frac{1}{x})}^3-3(x+\frac{1}{x})+{(x+\frac{1}{x})}^2-2+x+\frac{1}{x}=70$$$$letx+\frac{1}{x}=a$$$$a^3+a^2-2a-72=0$$$$(a-4)(a^2+5a+18)=0$$$$a=4,-\frac{5}{2}\pm \frac{\sqrt{47}i}{2}$$$$a=4$$$$so$$$$x+\frac{1}{x}=4$$$$x^2+\frac{1}{x^2}=14$$$$x^4+\frac{1}{x^4}=194$$

Commented by nurtani last updated on 25/Jun/22

$$\mathcal{T}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}!$$

Commented by Tawa11 last updated on 25/Jun/22

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

Commented by peter frank last updated on 27/Jun/22

$$\mathrm{thank}\mathrm{you}$$

Answered by greougoury555 last updated on 25/Jun/22

$$(x^2+x+1)(\frac{1}{x^2}+\frac{1}{x}+1)=70$$$$(x^2+x+1)\left(\frac{x^2+x+1}{x^2}\right)=70$$$${(x^2+x+1)}^2=70x^2$$$$(x^2+x+1-x\sqrt{70})(x^2+x+x\sqrt{70}+1)=0$$$$\{\begin{array}{l}{x}^2+(1-\sqrt{70})x+1=0\\ x^2+(1+\sqrt{70})x+1=0\end{array}$$$$$$

Answered by Rasheed.Sindhi last updated on 26/Jun/22

$$x^2(x+1+\frac{1}{x})+\frac{1}{x^2}(\frac{1}{x}+1+x)=70$$$$(x+1+\frac{1}{x})(x^2+\frac{1}{x^2})=70$$$$(x+\frac{1}{x}+1)({(x+\frac{1}{x})}^2-2)=70$$$$(y+1)(y^2-2)=70;x+\frac{1}{x}=y$$$$y^3+y^2-2y-72=$$$$(y-4)(y^2+5y+18)=0$$$$y=4\mid y=\frac{-5\pm \sqrt{25-72}}{2}=\frac{-5\pm i\sqrt{47}}{2}$$$$x+\frac{1}{x}=4$$$$x^2+\frac{1}{x^2}=16-2=14$$$$x^4+\frac{1}{x^4}=196-2=194$$

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