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Question Number 109611 by mnjuly1970 last updated on 24/Aug/20

Commented by Rasheed.Sindhi last updated on 25/Aug/20

${(x^{2} + \frac{1}{x^{2}})}^{3} = {(x^{2})}^{3} + \frac{1}{{(x^{2})}^{3}} + 3 (x^{2} + \frac{1}{x^{2}})$ ${(x^{2} + \frac{1}{x^{2}})}^{3} = {(x^{3})}^{2} + \frac{1}{{(x^{3})}^{2}} + 3 (x^{2} + \frac{1}{x^{2}})$ $= {(\sqrt{5} + 2)}^{2} + \frac{1}{{(\sqrt{5} + 2)}^{2}} + 3 (x^{2} + \frac{1}{x^{2}})$ $= 9 + 4 \sqrt{5} + \frac{1}{9 + 4 \sqrt{5}} + 3 (x^{2} + \frac{1}{x^{2}})$ $= 18 + 3 (x^{2} + \frac{1}{x^{2}})$ ${(x^{2} + \frac{1}{x^{2}})}^{3} - 3 (x^{2} + \frac{1}{x^{2}}) - 18 = 0$ $a^{3} - 3 a - 18 = 0; a = x^{2} + \frac{1}{x^{2}}$ $(a - 3) (a^{2} + 3 a + 6) = 0$ $a = 3$ $x^{2} + \frac{1}{x^{2}} = 3 (P r o v e d)$
	Answered by ajfour last updated on 25/Aug/20

	$I f x^{2} + \frac{1}{x^{2}} = 3$ ${(^{} x + \frac{1}{x})}^{2} = 5$ $(x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x}) = x^{3} + \frac{1}{x^{3}} + x + \frac{1}{x} = 3 \sqrt{5}$ $\Rightarrow x^{3} + \frac{1}{x^{3}} = 2 \sqrt{5} o r {(x^{3})}^{2} - 2 \sqrt{5} (x^{3}) + 1 = 0$ $\Rightarrow x^{3} = \sqrt{5} + \sqrt{5 - 1} = \sqrt{5} + 2$ $h e n c e x^{2} + \frac{1}{x^{2}} = 3 (I n d e e d!)$
	Answered by Her_Majesty last updated on 24/Aug/20

	$x^{3} = 2 + \sqrt{5} \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{2} \land x_{2} = ω x_{1} \land x_{3} = ω^{2} x_{1}$ $\Rightarrow x_{1}^{2} = \frac{3 + \sqrt{5}}{2} \land \frac{1}{x_{1}^{2}} = \frac{3 - \sqrt{5}}{2} \land x_{2}^{2} e t c e a s y t o g e t$ $\Rightarrow p r o v e n$ $t o l a z y t o w r i t e i t o u t$
	Answered by Rasheed.Sindhi last updated on 24/Aug/20

	${(x^{3})}^{2} = {(2 + \sqrt{5})}^{2} = 9 + 4 \sqrt{5}$ $x^{6} = 9 + 4 \sqrt{5}, \frac{1}{x^{6}} = 9 - 4 \sqrt{5}$ $x^{6} + \frac{1}{x^{6}} = (9 + 4 \sqrt{5}) + (9 - 4 \sqrt{5})$ $x^{6} + \frac{1}{x^{6}} = 18$ $N o w,$ ${(x^{2} + \frac{1}{x^{2}})}^{3} = x^{6} + \frac{1}{x^{6}} + 3 (x^{2} + \frac{1}{x^{2}})$ $= 18 + 3 (x^{2} + \frac{1}{x^{2}})$ ${(x^{2} + \frac{1}{x^{2}})}^{3} - 3 (x^{2} + \frac{1}{x^{2}}) - 18 = 0$ $x^{2} + \frac{1}{x^{2}} = u$ $u^{3} - 3 u - 18 = 0$ $(u - 3) (u^{2} + 3 u + 6) = 0$ $u = 3$ $O r x^{2} + \frac{1}{x^{2}} = 3 p r o v e d$
	Commented by mnjuly1970 last updated on 24/Aug/20

	$t h a n k y o u s i r . .$
	Answered by 1549442205PVT last updated on 25/Aug/20
	$x^3 =2+(√5) ⇒x=^3 (√(2+(√5))) ,(1/x)=(1/(^3 (√(2+(√5))) )) Note (2+(√5))(2−(√5))=2^2 −((√5))^2 =−1 ⇒ (1/x)=(1/(^3 (√(2+(√5))) ))=((^3 (√(2−(√5))) )/(^3 ((√(2+(√5))) )(^3 (√(2+(√5))) ))) =^3 (√(2−(√5))) .From that x^2 =(^3 (√(2+(√5))) )^2 =^3 (√(9+4(√5))) , (1/x^2 )=(^3 (√(2−(√5))) )^2 =^3 (√(9−4(√5))) (∗) Put 9+4(√5)=(x+y(√5))^3 =x^3 +3x^2 y(√5) +15xy^2 +5y^3 (√5)=(x^3 +15xy^2 )+(3x^2 y+5y^3 )(√5) ⇔ { ((x^3 +15xy^2 =9(1))),((3x^2 y+5y^3 =4(2))) :} ⇒4x^3 +60xy^2 =27x^2 y+45y^3 .Put x=ky we get 4k^3 −27k^2 +60k−45=0 ⇔(k−3)(4k^2 −15k+15)=0⇒k=3 ⇒x=3y.Replace into (1)we get 27y^3 +45y^3 =9⇔72y^3 =9⇔y^3 =(1/8) ⇒y=(1/2),x=(3/2).Hence x^2 =^3 (√(9+4(√5))) =^3 (√(((3/2)+((√5)/2))))^3 =(3/2)+((√5)/2) Similarly,(1/x^2 )=^3 (√(9−4(√5)))=(3/2)−((√5)/2). ⇒x^2 +(1/x^2 )=3 (Q.E.D)$
	$x^{3} = 2 + \sqrt{5} \Rightarrow x =^{3} \sqrt{2 + \sqrt{5}}, \frac{1}{x} = \frac{1}{^{3} \sqrt{2 + \sqrt{5}}}$ $Note (2 + \sqrt{5}) (2 - \sqrt{5}) = 2^{2} - {(\sqrt{5})}^{2} = - 1$ $\Rightarrow \frac{1}{x} = \frac{1}{^{3} \sqrt{2 + \sqrt{5}}} = \frac{^{3} \sqrt{2 - \sqrt{5}}}{^{3} (\sqrt{2 + \sqrt{5}}) (^{3} \sqrt{2 + \sqrt{5}})}$ $=^{3} \sqrt{2 - \sqrt{5}} . From that$ $x^{2} = {(^{3} \sqrt{2 + \sqrt{5}})}^{2} =^{3} \sqrt{9 + 4 \sqrt{5}},$ $\frac{1}{x^{2}} = {(^{3} \sqrt{2 - \sqrt{5}})}^{2} =^{3} \sqrt{9 - 4 \sqrt{5}} (*)$ $Put 9 + 4 \sqrt{5} = {(x + y \sqrt{5})}^{3} = x^{3} + {3 x}^{2} y \sqrt{5}$ $+ {15 xy}^{2} + {5 y}^{3} \sqrt{5} = (x^{3} + {15 xy}^{2}) + ({3 x}^{2} y + {5 y}^{3}) \sqrt{5}$ $\Leftrightarrow {\begin{cases} x^{3} + {15 xy}^{2} = 9 (1) \\ {3 x}^{2} y + {5 y}^{3} = 4 (2) \end{cases}$ $\Rightarrow {4 x}^{3} + {60 xy}^{2} = {27 x}^{2} y + {45 y}^{3} . Put x = ky$ $we get {4 k}^{3} - {27 k}^{2} + 60 k - 45 = 0$ $\Leftrightarrow (k - 3) ({4 k}^{2} - 15 k + 15) = 0 \Rightarrow k = 3$ $\Rightarrow x = 3 y . Replace into (1) we get$ ${27 y}^{3} + {45 y}^{3} = 9 \Leftrightarrow {72 y}^{3} = 9 \Leftrightarrow y^{3} = \frac{1}{8}$ $\Rightarrow y = \frac{1}{2}, x = \frac{3}{2} . Hence$ ${x^{2} =^{3} \sqrt{9 + 4 \sqrt{5}} =^{3} \sqrt{(\frac{3}{2} + \frac{\sqrt{5}}{2}})}^{3} = \frac{3}{2} + \frac{\sqrt{5}}{2}$ $Similarly, \frac{1}{x^{2}} =^{3} \sqrt{9 - 4 \sqrt{5}} = \frac{3}{2} - \frac{\sqrt{5}}{2} .$ $\Rightarrow x^{2} + \frac{1}{x^{2}} = 3 (Q . E . D)$
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