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Question Number 200961 by jabarsing last updated on 27/Nov/23

−x^3 +1=((−x+1))^(1/3) ⇒x=?

$$\: \\ $$$$\:\:\:\:\:−{x}^{\mathrm{3}} +\mathrm{1}=\sqrt[{\mathrm{3}}]{−{x}+\mathrm{1}}\:\Rightarrow{x}=? \\ $$$$ \\ $$

Answered by cortano12 last updated on 27/Nov/23

Answered by Frix last updated on 27/Nov/23

1−x^3 =(1−x)^(1/3) (1−x^3 )^3 =1−x x^9 −3x^6 +3x^3 −x=0 x(x−1)(x^3 +x−1)(x^4 +x^3 −2x−1)=0 x=0 x=1 x=(((1/2)+((√(93))/(18))))^(1/3) −((−(1/2)+((√(93))/(18))))^(1/3) ≈.682327804 x=((−1+(√5)−(√(−2+6(√5))))/4)≈−.535687387 x=((−1+(√5)+(√(−2+6(√5))))/4)≈1.15372138

$$\mathrm{1}−{x}^{\mathrm{3}} =\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\mathrm{3}} =\mathrm{1}−{x} \\ $$$${x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{3}} −{x}=\mathrm{0} \\ $$$${x}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}}\approx.\mathrm{682327804} \\ $$$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}−\sqrt{−\mathrm{2}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{4}}\approx−.\mathrm{535687387} \\ $$$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}+\sqrt{−\mathrm{2}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{15372138} \\ $$

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