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Question Number 185972 by Michaelfaraday last updated on 30/Jan/23

Answered by Rasheed.Sindhi last updated on 30/Jan/23

x=(p)^(1/3) +(1/( (p)^(1/3) )) , y=(√p) +(1/( (√p) )) Show that y^2 −2=x(x^2 −3) x^3 =((p)^(1/3) +(1/( (p)^(1/3) )))^3 =p+(1/p)+3((p)^(1/3) +(1/( (p)^(1/3) ))) =p+(1/p)+3(x) p+(1/p)=x^3 −3x=x(x^2 −3)......(i) y^2 =((√p) +(1/( (√p) )))^2 =p+(1/p)+2 p+(1/p)=y^2 −2.............(ii) (i) & (ii): y^2 −2=x(x^2 −3)

$${x}=\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\:,\:{y}=\sqrt{{p}}\:+\frac{\mathrm{1}}{\:\sqrt{{p}}\:} \\ $$$${Show}\:{that}\:\:\:{y}^{\mathrm{2}} −\mathrm{2}={x}\left({x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$${x}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right)^{\mathrm{3}} ={p}+\frac{\mathrm{1}}{{p}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right) \\ $$$$\:\:\:\:={p}+\frac{\mathrm{1}}{{p}}+\mathrm{3}\left({x}\right) \\ $$$${p}+\frac{\mathrm{1}}{{p}}={x}^{\mathrm{3}} −\mathrm{3}{x}={x}\left({x}^{\mathrm{2}} −\mathrm{3}\right)......\left({i}\right) \\ $$$${y}^{\mathrm{2}} =\left(\sqrt{{p}}\:+\frac{\mathrm{1}}{\:\sqrt{{p}}\:}\right)^{\mathrm{2}} ={p}+\frac{\mathrm{1}}{{p}}+\mathrm{2} \\ $$$${p}+\frac{\mathrm{1}}{{p}}={y}^{\mathrm{2}} −\mathrm{2}.............\left({ii}\right) \\ $$$$\left({i}\right)\:\&\:\left({ii}\right):\:\:\:{y}^{\mathrm{2}} −\mathrm{2}={x}\left({x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$

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