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Solve-for-x-x-3-3x-3-1-3-x-3-3x-5-1-3-2-An-alteration-of-Q-202500-

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Question Number 202535 by Rasheed.Sindhi last updated on 29/Dec/23
Solve for x ((x^3 +3x−3))^(1/3) +((−x^3 −3x+5))^(1/3) =2 (An alteration of Q#202500)
$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$
Answered by mr W last updated on 29/Dec/23
((x^3 +3x−3))^(1/n) +((−x^3 −3x+5))^(1/n) =2 n∈N^(≥2) , x∈R let x^3 +3x−4=u ((1+u))^(1/n) +((1−u))^(1/n) =2 u=0 is a trivial root ((1+u))^(1/n) is strictly increasing. ((1−u))^(1/n) is strictly decreasing. ⇒u=0 is one and the only one root ⇒x^3 +3x−4=u=0 ⇒(x−1)(x^2 +x+4)=0 x^2 +x+4=(x+(1/2))^2 +((15)/4)≠0 ⇒x−1=0 ⇒x=1 ✓ or ((1+u))^(1/n) +((1−u))^(1/n) ≤2 equality holds when ((1+u))^(1/n) =((1−u))^(1/n) , i.e. 1+u=1−u, ⇒u=0
$$\sqrt[{{n}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}+\sqrt[{{n}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}=\mathrm{2} \\ $$$${n}\in{N}^{\geqslant\mathrm{2}} ,\:\boldsymbol{{x}}\in\boldsymbol{{R}} \\ $$$${let}\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}={u} \\ $$$$\sqrt[{{n}}]{\mathrm{1}+{u}}+\sqrt[{{n}}]{\mathrm{1}−{u}}=\mathrm{2} \\ $$$${u}=\mathrm{0}\:{is}\:{a}\:{trivial}\:{root} \\ $$$$\sqrt[{{n}}]{\mathrm{1}+{u}}\:{is}\:{strictly}\:{increasing}. \\ $$$$\sqrt[{{n}}]{\mathrm{1}−{u}}\:{is}\:{strictly}\:{decreasing}. \\ $$$$\Rightarrow{u}=\mathrm{0}\:{is}\:{one}\:{and}\:{the}\:{only}\:{one}\:{root} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}={u}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{4}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{4}}\neq\mathrm{0} \\ $$$$\Rightarrow{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1}\:\checkmark \\ $$$$ \\ $$$${or}\: \\ $$$$\sqrt[{{n}}]{\mathrm{1}+{u}}+\sqrt[{{n}}]{\mathrm{1}−{u}}\leqslant\mathrm{2} \\ $$$${equality}\:{holds}\:{when}\:\sqrt[{{n}}]{\mathrm{1}+{u}}=\sqrt[{{n}}]{\mathrm{1}−{u}}, \\ $$$${i}.{e}.\:\mathrm{1}+{u}=\mathrm{1}−{u},\:\Rightarrow{u}=\mathrm{0} \\ $$
Commented by mr W last updated on 29/Dec/23
in case n=3: ((1+u))^(1/3) +((1−u))^(1/3) =a with 0<a≤2 1+u+1−u+3(((1+u)(1−u)))^(1/3) (((1+u))^(1/3) +((1−u))^(1/3) )=2^3 2+3a(((1+u)(1−u)))^(1/3) =a^3 ((1−u^2 ))^(1/3) =((a^3 −2)/(3a)) ⇒1−u^2 =(((a^3 −2)^3 )/(27a^3 )) ⇒u=±(√(1−(((a^3 −2)^3 )/(27a^3 )))) ...
$${in}\:{case}\:{n}=\mathrm{3}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{1}+{u}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−{u}}={a}\:{with}\:\mathrm{0}<{a}\leqslant\mathrm{2} \\ $$$$\mathrm{1}+{u}+\mathrm{1}−{u}+\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+{u}\right)\left(\mathrm{1}−{u}\right)}\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+{u}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−{u}}\right)=\mathrm{2}^{\mathrm{3}} \\ $$$$\mathrm{2}+\mathrm{3}{a}\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+{u}\right)\left(\mathrm{1}−{u}\right)}={a}^{\mathrm{3}} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{1}−{u}^{\mathrm{2}} }=\frac{{a}^{\mathrm{3}} −\mathrm{2}}{\mathrm{3}{a}} \\ $$$$\Rightarrow\mathrm{1}−{u}^{\mathrm{2}} =\frac{\left({a}^{\mathrm{3}} −\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} } \\ $$$$\Rightarrow{u}=\pm\sqrt{\mathrm{1}−\frac{\left({a}^{\mathrm{3}} −\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }} \\ $$$$… \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/23
Thanks sir mr W!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mr}}\:\boldsymbol{\mathrm{W}}! \\ $$

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