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Question Number 96650 by bobhans last updated on 03/Jun/20

$$\mathrm{solve}2\sqrt[3]{2\mathrm{y}-1}={\mathrm{y}}^3+1$$

Answered by john santu last updated on 03/Jun/20

$$\sqrt[3]{2\mathrm{y}-1}=\frac{{\mathrm{y}}^3+1}{2}$$$$\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{y}\right)\Rightarrow \mathrm{y}=\frac{{\mathrm{y}}^3+1}{2}$$$${\mathrm{y}}^3-2\mathrm{y}+1=0;(\mathrm{y}-1)({\mathrm{y}}^2+\mathrm{y}-1)=0$$$$\{\begin{array}{l}\mathrm{y}=1\\ \mathrm{y}=\frac{-1\pm \sqrt{5}}{2}\end{array}$$

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