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Question-165419




Question Number 165419 by mathlove last updated on 01/Feb/22
Answered by TheSupreme last updated on 01/Feb/22
(a+(1/a))=x  a^2 +(1/a^2 )=(a+(1/a))^2 −2=x^2 −2  a^3 +(1/a^3 )=(a+(1/a))^3 −3a−3(1/a)=x^3 −3x  x+x^2 −2+x^3 −3x=28  x^3 +x^2 −2x−30=0  (x−3)(x^2 +4x+10)=0  x=3  a+(1/a)=3  a^2 −3a+1=0  a=((3±(√5))/2)  (2a−3)^2 =(±(√5))^2 =5
$$\left({a}+\frac{\mathrm{1}}{{a}}\right)={x} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}={x}^{\mathrm{2}} −\mathrm{2} \\ $$$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} −\mathrm{3}{a}−\mathrm{3}\frac{\mathrm{1}}{{a}}={x}^{\mathrm{3}} −\mathrm{3}{x} \\ $$$${x}+{x}^{\mathrm{2}} −\mathrm{2}+{x}^{\mathrm{3}} −\mathrm{3}{x}=\mathrm{28} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{30}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{3} \\ $$$${a}^{\mathrm{2}} −\mathrm{3}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} =\left(\pm\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{5} \\ $$

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