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Question-153505




Question Number 153505 by mathdanisur last updated on 07/Sep/21
Commented by MJS_new last updated on 08/Sep/21
((x^2 +y^2 +5))^(1/3) −(√(x^2 +y^2 −5))=5 has no solution ∈C
$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{5}}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{5}}=\mathrm{5}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\in\mathbb{C} \\ $$
Answered by liberty last updated on 08/Sep/21
 let x^2 +y^2 =u  ⇒((u+5))^(1/3) −(√(u−5)) = 5  let ((u+5))^(1/3)  =v⇒u=v^3 −5  ⇒v−5=(√(v^3 −10))  ⇒v^2 −10v+25=v^3 −10  ⇒v^3 −v^2 +10v−35=0    Let λ=((−25+x^4 +y^4 +2x^2 y^2 ))^(1/3)   λ=(((x^2 +y^2 )^2 −25))^(1/3) =((u^2 −25))^(1/3)   λ=(((u−5)(u+5)))^(1/3)
$$\:{let}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={u} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{{u}+\mathrm{5}}−\sqrt{{u}−\mathrm{5}}\:=\:\mathrm{5} \\ $$$${let}\:\sqrt[{\mathrm{3}}]{{u}+\mathrm{5}}\:={v}\Rightarrow{u}={v}^{\mathrm{3}} −\mathrm{5} \\ $$$$\Rightarrow{v}−\mathrm{5}=\sqrt{{v}^{\mathrm{3}} −\mathrm{10}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} −\mathrm{10}{v}+\mathrm{25}={v}^{\mathrm{3}} −\mathrm{10} \\ $$$$\Rightarrow{v}^{\mathrm{3}} −{v}^{\mathrm{2}} +\mathrm{10}{v}−\mathrm{35}=\mathrm{0} \\ $$$$ \\ $$$${Let}\:\lambda=\sqrt[{\mathrm{3}}]{−\mathrm{25}+{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\lambda=\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{25}}=\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\lambda=\sqrt[{\mathrm{3}}]{\left({u}−\mathrm{5}\right)\left({u}+\mathrm{5}\right)}\: \\ $$$$ \\ $$
Commented by mr W last updated on 08/Sep/21
((u+5))^(1/3) −(√(u−5)) = 5 has no solution!
$$\sqrt[{\mathrm{3}}]{{u}+\mathrm{5}}−\sqrt{{u}−\mathrm{5}}\:=\:\mathrm{5}\:{has}\:{no}\:{solution}! \\ $$

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