((▽BeMath▽)/△) If x^4 +x^2 = ((11)/5) , find the value of Ω = (((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3)


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Question Number 108316 by bemath last updated on 16/Aug/20

$\frac{▽ B e M a t h ▽}{△}$ $I f x^{4} + x^{2} = \frac{11}{5}, f i n d t h e v a l u e o f$ $Ω = \sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}}$
	Answered by bobhans last updated on 16/Aug/20
	$((bobhans)/𝛙) consider (x)^(1/3) +(y)^(1/3) = ((x+y)/( (x^2 )^(1/3) −((xy))^(1/3) +(y^2 )^(1/3) )) now (((x+1)/(x−1)))^(1/3) +(((x−1)/(x+1)))^(1/3) = ((((x+1)/(x−1))+((x−1)/(x+1)))/( (((((x+1)/(x−1)))^2 ))^(1/3) +(((((x−1)/(x+1)))^2 ))^(1/3) −1)) =((2x^2 +2)/((x^2 −1){(((((x+1)/(x−1)))^2 ))^(1/3) +(((((x−1)/(x+1)))^2 ))^(1/3) −1})) (∗)from the equation x^4 +x^2 =((11)/5) let x^2 =z ⇒5z^2 +5z−11=0 z= ((−5 ± (√(25+220)))/(10)) = ((−5 ± 7(√5))/(10)) if x∈R , we take z = ((−5 + 7(√5))/(10)) ≈ 1.065 and x = (√(1.065)) ≈ 1.032$
	$\frac{b o b h a n s}{ψ}$ $c o n s i d e r \sqrt[3]{x} + \sqrt[3]{y} = \frac{x + y}{\sqrt[3]{x^{2}} - \sqrt[3]{x y} + \sqrt[3]{y^{2}}}$ $n o w \sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}} = \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x + 1}}{\sqrt[3]{{(\frac{x + 1}{x - 1})}^{2}} + \sqrt[3]{{(\frac{x - 1}{x + 1})}^{2}} - 1}$ $= \frac{2 x^{2} + 2}{(x^{2} - 1) {\sqrt[3]{{(\frac{x + 1}{x - 1})}^{2}} + \sqrt[3]{{(\frac{x - 1}{x + 1})}^{2}} - 1}}$ $(*) f r o m t h e e q u a t i o n x^{4} + x^{2} = \frac{11}{5}$ $l e t x^{2} = z \Rightarrow 5 z^{2} + 5 z - 11 = 0$ $z = \frac{- 5 \pm \sqrt{25 + 220}}{10} = \frac{- 5 \pm 7 \sqrt{5}}{10}$ $i f x \in R, w e t a k e z = \frac{- 5 + 7 \sqrt{5}}{10} \approx 1.065$ $a n d x = \sqrt{1.065} \approx 1.032$
	Answered by Rasheed.Sindhi last updated on 16/Aug/20

	$x^{4} + x^{2} = \frac{11}{5} \Rightarrow x^{2} = \frac{- 5 \pm 7 \sqrt{5}}{10}$ $Ω = \sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}}$ $Ω^{3} = {(\sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}})}^{3}$ $= \frac{x + 1}{x - 1} + \frac{x - 1}{x + 1} + 3 (\sqrt[3]{\frac{x + 1}{x - 1}} + \sqrt[3]{\frac{x - 1}{x + 1}})$ $Ω^{3} - 3 Ω = \frac{2 (x^{2} + 1)}{x^{2} - 1}$ $= 2 (\frac{\frac{- 5 \pm 7 \sqrt{5}}{10} + 1}{\frac{- 5 \pm 7 \sqrt{5}}{10} - 1}) = 2 (\frac{5 \pm 7 \sqrt{5}}{- 15 \pm 7 \sqrt{5}})$ $= 2 (\frac{5 \pm 7 \sqrt{5}}{- 15 \pm 7 \sqrt{5}} \times \frac{- 15 \mp 7 \sqrt{5}}{- 15 \mp 7 \sqrt{5}})$ $= 2 (\frac{- 75 \mp 35 \sqrt{5} \mp 105 \sqrt{5} - 49 (5)}{225 - 49 (5)})$ $= \frac{320 \pm 140 \sqrt{5}}{10} = 32 \pm 14 \sqrt{5}$
	Commented by bemath last updated on 16/Aug/20

	$c o l l a n d n i c e$
	Answered by mr W last updated on 16/Aug/20

	$x^{4} + x^{2} - \frac{11}{5} = 0$ $x^{2} = \frac{1}{2} (- 1 + \sqrt{1 + \frac{44}{5}}) = \frac{7 \sqrt{5} - 5}{10} > 0$ $\frac{x^{2} + 1}{x^{2} - 1} = \frac{7 \sqrt{5} + 5}{7 \sqrt{5} - 15} = 16 + 7 \sqrt{5}$ ${(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)$ $l e t a = \sqrt[3]{\frac{x + 1}{x - 1}}, b = \sqrt[3]{\frac{x - 1}{x + 1}}$ $Ω^{3} = 2 \times \frac{x^{2} + 1}{x^{2} - 1} + 3 Ω$ $\Rightarrow Ω^{3} - 3 Ω - 2 λ = 0 w i t h λ = \frac{x^{2} + 1}{x^{2} - 1} = 16 + 7 \sqrt{5}$ $Δ = {(- 1)}^{3} + {(- λ)}^{2} = 4 (125 + 56 \sqrt{5}) > 0$ $Ω = \sqrt[3]{\sqrt{Δ} + λ} - \sqrt[3]{\sqrt{Δ} - λ}$ $= \sqrt[3]{2 \sqrt{125 + 56 \sqrt{5}} + 16 + 7 \sqrt{5}} - \sqrt[3]{2 \sqrt{125 + 56 \sqrt{7}} - 16 - 7 \sqrt{5}}$ $\approx 4.236068$
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