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Question Number 216694 by sniper237 last updated on 16/Feb/25
Prove that ^3 (√((√5)+2)) −^3 (√((√5)−2)) =1
$${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$
Answered by golsendro last updated on 16/Feb/25
let x= (((√5)+2))^(1/3) −(((√5)−2))^(1/3) (((√5)+2))^(1/3) −(((√5)−2))^(1/3) −x = 0 (√5) +2 −(√5) +2 −x^3 = 3x ((5−4))^(1/3) x^3 +3x−4 = 0 x=1
$$\:\mathrm{let}\:\mathrm{x}=\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}}−\mathrm{x}\:=\:\mathrm{0}\: \\ $$$$\:\sqrt{\mathrm{5}}\:+\mathrm{2}\:−\sqrt{\mathrm{5}}\:+\mathrm{2}\:−\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3x}\:\sqrt[{\mathrm{3}}]{\mathrm{5}−\mathrm{4}} \\ $$$$\:\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}−\mathrm{4}\:=\:\mathrm{0}\: \\ $$$$\:\:\mathrm{x}=\mathrm{1}\: \\ $$
Answered by Ghisom last updated on 16/Feb/25
Φ=(1/2)+((√5)/2) Φ^(−1) =−(1/2)+((√5)/2) Φ^3 =2+(√5) Φ^(−3) =−2+(√5) the rest is easy
$$\Phi=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Phi^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Phi^{\mathrm{3}} =\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\Phi^{−\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Feb/25
(((√5) +2))^(1/3) −(((√5) −2))^(1/3) =1 let (((√5) +2))^(1/3) =a (((√5) −2))^(1/3) =1/a Asume a−(1/a)=x a^3 −(1/a^3 )−3(a−(1/a))=x^3 ((√5) +2)−((√5) −2)−3(x)=x^3 x^3 +3x−4=0 (x−1)(x^2 +x+4)=0 ⇒x=1
$$\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1} \\ $$$${let}\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:={a} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1}/{a} \\ $$$${Asume}\:{a}−\frac{\mathrm{1}}{{a}}={x} \\ $$$${a}^{\mathrm{3}} −\frac{\mathrm{1}}{{a}^{\mathrm{3}} }−\mathrm{3}\left({a}−\frac{\mathrm{1}}{{a}}\right)={x}^{\mathrm{3}} \\ $$$$\left(\sqrt{\mathrm{5}}\:+\mathrm{2}\right)−\left(\sqrt{\mathrm{5}}\:−\mathrm{2}\right)−\mathrm{3}\left({x}\right)={x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Feb/25
(((√5) +2))^(1/3) −(((√5) −2))^(1/3) =1 let (((√5) +2))^(1/3) −(((√5) −2))^(1/3) =x (((√5) +2))^(1/3) −(((√5) −2))^(1/3) −x=0 −((−(√5) −2))^(1/3) −(((√5) −2))^(1/3) −x=0 ((−(√5) −2))^(1/3) +(((√5) −2))^(1/3) +x=0 determinant (((a+b+c=0_(⇒a^3 +b^3 +c^3 =3abc) ))) ⇒(−(√5) −2)+((√5) −2)+x^3 =3( ((−(√5) −2))^(1/3) )((((√5) −2))^(1/3) )(x) x^3 −4=−3x((((√5) +2)((√5)−2)))^(1/3) x^3 +3x(1)−4=0 (x−1)(x^2 +x+4)=0 ⇒x=1(proved)
$$\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1} \\ $$$${let}\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:={x} \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:−{x}=\mathrm{0} \\ $$$$\:\:\:\:−\sqrt[{\mathrm{3}}]{−\sqrt{\mathrm{5}}\:−\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:−{x}=\mathrm{0} \\ $$$$\:\:\:\:\sqrt[{\mathrm{3}}]{−\sqrt{\mathrm{5}}\:−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:+{x}=\mathrm{0} \\ $$$$\begin{array}{|c|}{\underset{\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}} {{a}+{b}+{c}=\mathrm{0}}}\\\hline\end{array} \\ $$$$\Rightarrow\left(−\sqrt{\mathrm{5}}\:−\mathrm{2}\right)+\left(\sqrt{\mathrm{5}}\:−\mathrm{2}\right)+{x}^{\mathrm{3}} \\ $$$$\:\:\:=\mathrm{3}\left(\:\sqrt[{\mathrm{3}}]{−\sqrt{\mathrm{5}}\:−\mathrm{2}}\right)\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\right)\left({x}\right) \\ $$$${x}^{\mathrm{3}} −\mathrm{4}=−\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{5}}\:+\mathrm{2}\right)\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}\: \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}\left(\mathrm{1}\right)−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\left({proved}\right) \\ $$

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