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Question Number 68768 by aliesam last updated on 15/Sep/19

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

Commented by kaivan.ahmadi last updated on 15/Sep/19

$$t=x-1\Rightarrow x=t+1\Rightarrow$$$$\sqrt[3]{2t+1}+\sqrt[3]{t}=1\Rightarrow \sqrt[3]{2t+1}=1-\sqrt[3]{t}\Rightarrow$$$$2t+1=1-3\sqrt[3]{t}+3\sqrt[3]{t^2}-t\Rightarrow 3t=3\sqrt[3]{t}(\sqrt[3]{t}-1)\Rightarrow$$$$(let\sqrt[3]{t}=0\Rightarrow t=0\Rightarrow x=1)otherwise$$$$\sqrt[3]{t^2}=\sqrt[3]{t}-1\Rightarrow \sqrt[3]{t^2}-\sqrt[3]{t}+1=0$$$$sety=\sqrt[3]{t}\Rightarrow y^2-y+1=0\Rightarrow$$$$y_{1,2}=\frac{1\pm \sqrt{3}i}{2}$$$$\{\begin{array}{l}y=\frac{1+\sqrt{3}i}{2}\Rightarrow t={\left(\frac{1+\sqrt{3}i}{2}\right)}^3\Rightarrow x={\left(\frac{1+\sqrt{3}i}{2}\right)}^3+1\\ y=\frac{1-\sqrt{3}i}{2}\Rightarrow t={\left(\frac{1-\sqrt{3}i}{2}\right)}^3\Rightarrow x={\left(\frac{1-\sqrt{3}i}{2}\right)}^3+1\end{array}$$$$$$

Answered by MJS last updated on 16/Sep/19

$$\mathrm{obviously}x=1\mathrm{is}\mathrm{a}\mathrm{solution}$$$$a^{1/3}+b^{1/3}=c^{1/3}$$$$a+3a^{1/3}{b}^{1/3}(a^{1/3}+b^{1/3})+b=c$$$$a^{1/3}{b}^{1/3}{c}^{1/3}=c-a-b$$$$abc={(c-a-b)}^3$$$$27(2x-1)(x-1)={(3-3x)}^3$$$$(2x-1)(x-1)={(1-x)}^3$$$$x^2(x-1)=0\Rightarrow x=1(x=0\mathrm{not}\mathrm{valid})$$

Commented by Rasheed.Sindhi last updated on 16/Sep/19

$$\mathbf{Sir}4\mathrm{th}\mathrm{line}:a^{1/3}{b}^{1/3}{c}^{1/3}=c-a-b$$$$$$

Commented by MJS last updated on 16/Sep/19

$$\mathrm{thank}\mathrm{you},\mathrm{I}\mathrm{corrected}\mathrm{it}$$

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