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Question-89795




Question Number 89795 by A8;15: last updated on 19/Apr/20
Answered by MJS last updated on 19/Apr/20
(x^2 +1)(x+1)^2 +x^2 =(x^2 +x+1)^2   x^2 (x^2 +1)+1=(x^2 −x+1)(x^2 +x+1)  ⇒  ((x^2 +x+1)/(x^2 −x+1))=x+(1/x)  x^4 −2x^3 +x^2 −2x+1=0  (x^2 −(√2)x−x+1)(x^2 +(√2)x−x+1)=0  and this is easy to solve
$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{1}=\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$\Rightarrow \\ $$$$\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}={x}+\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}−{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}−{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

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