Question-223405

Question Number 223405 by behi834171 last updated on 24/Jul/25

Answered by fantastic last updated on 24/Jul/25

((x+((x+1))^(1/3) ))^(1/3) =(1/2) (((x+((x+1))^(1/3) ))^(1/3) )^3 =(1/8) ∴x+((x+1))^(1/3) =(1/8) let ((x+1))^(1/3) =u So u^3 =x+1 or x=u^3 −1 So the equation becomes u^3 −1+u=(1/8) u≈0.73229785062139 So x=u^3 −1⇒≈−0.60729785062139

$$\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\therefore{x}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${let}\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}={u} \\ $$$${So}\:{u}^{\mathrm{3}} ={x}+\mathrm{1} \\ $$$${or}\:{x}={u}^{\mathrm{3}} −\mathrm{1} \\ $$$${So}\:{the}\:{equation}\:{becomes} \\ $$$${u}^{\mathrm{3}} −\mathrm{1}+{u}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${u}\approx\mathrm{0}.\mathrm{73229785062139} \\ $$$${So}\:{x}={u}^{\mathrm{3}} −\mathrm{1}\Rightarrow\approx−\mathrm{0}.\mathrm{60729785062139} \\ $$

Answered by mr W last updated on 24/Jul/25

x+1+((x+1))^(1/3) =(9/8) t=((x+1))^(1/3) t^3 +t−(9/8)=0 t=(((√((−(9/(16)))^2 +(1/3^3 )))+(9/(16))))^(1/3) −(((√((−(9/(16)))^2 +(1/3^3 )))−(9/(16))))^(1/3) =((((√(7329))/(144))+(9/(16))))^(1/3) −((((√(7329))/(144))−(9/(16))))^(1/3) x=t^3 −1=(1/8)−t =(1/8)−((((√(7329))/(144))+(9/(16))))^(1/3) +((((√(7329))/(144))−(9/(16))))^(1/3) ≈−0.607298

$${x}+\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}=\frac{\mathrm{9}}{\mathrm{8}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}} \\ $$$${t}^{\mathrm{3}} +{t}−\frac{\mathrm{9}}{\mathrm{8}}=\mathrm{0} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }}+\frac{\mathrm{9}}{\mathrm{16}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{16}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }}−\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$$\:=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7329}}}{\mathrm{144}}+\frac{\mathrm{9}}{\mathrm{16}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7329}}}{\mathrm{144}}−\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$${x}={t}^{\mathrm{3}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{8}}−{t} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{8}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7329}}}{\mathrm{144}}+\frac{\mathrm{9}}{\mathrm{16}}}+\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{7329}}}{\mathrm{144}}−\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$$\:\:\approx−\mathrm{0}.\mathrm{607298} \\ $$

Commented by fantastic last updated on 24/Jul/25

$${I}\:{think}\:{you}\:{used}\:{Cardano}'{s}\:{Formula}.\:{Right}?? \\ $$

Commented by mr W last updated on 24/Jul/25

$${yes} \\ $$

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