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Question Number 167508 by Bagus1003 last updated on 18/Mar/22

$$2\sqrt[3]{2x+1}=x^3-1$$$$Howmuchthexis?$$

Answered by alephzero last updated on 18/Mar/22

$$2\sqrt[3]{2x+1}=x^3-1$$$$\Rightarrow \sqrt[3]{2x+1}=\frac{x^3-1}{2}$$$$\Rightarrow 2x+1=\frac{x^9-3x^6+3x^3-1}{8}$$$$\Rightarrow 2x=\frac{x^9-3x^6+3x^3-2}{8}$$$$\Rightarrow x=\frac{x^9-3x^6+3x^3-2}{16}$$$$\Rightarrow \frac{x^9-3x^6+3x^3-8x-2}{16}=0$$$$\Rightarrow x^9-3x^6+3x^3-8x-2=0$$$$x_1\approx -0.9592$$$$x_2\approx -0.2564$$$$x_3\approx 1.4962$$

Answered by Jamshidbek last updated on 18/Mar/22

$$\mathrm{Solution}.$$$$2\mathrm{x}+1=\mathrm{t}\Rightarrow 2\mathrm{x}=\mathrm{t}-1$$$$2\sqrt[3]{2\mathrm{x}+1}={\mathrm{x}}^3-1\Rightarrow 16\sqrt[3]{2\mathrm{x}+1}={\left(2\mathrm{x}\right)}^3-8\Rightarrow$$$$16\mathrm{t}={(\mathrm{t}-1)}^3-8\Rightarrow {\mathrm{t}}^3-{3\mathrm{t}}^2+3\mathrm{t}-1-8=16\mathrm{t}$$$${\mathrm{t}}^3-{3\mathrm{t}}^2-13\mathrm{t}-9=0$$$${\mathrm{t}}^3+{\mathrm{t}}^2-{4\mathrm{t}}^2-4\mathrm{t}-9\mathrm{t}-9=0$$$${\mathrm{t}}^2(\mathrm{t}+1)-4\mathrm{t}(\mathrm{t}+1)-9(\mathrm{t}+1)=0$$$$(\mathrm{t}+1)({\mathrm{t}}^2-4\mathrm{t}-9)=0$$$${\mathrm{t}}_1=-1{\mathrm{t}}_{2,3}=\frac{4\pm \sqrt{52}}{2}=2\pm \sqrt{13}$$$${\mathrm{x}}_1=-1{\mathrm{x}}_{2,3}=\frac{2\pm \sqrt{13}-1}{2}=\frac{1\pm \sqrt{13}}{2}$$$$\mathrm{Telegram}:@\mathrm{math}\mathrm{\_}\mathrm{undergraduate}$$

Commented by mr W last updated on 18/Mar/22

$${\mathrm{x}}_1=-1{\mathrm{x}}_{2,3}=\frac{1\pm \sqrt{5}}{2}$$

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