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

solve 2 ((2y−1))^(1/(3  ))  = y^3 +1

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

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

((2y−1))^(1/(3  ))  = ((y^3 +1)/2)  ⇒f^(−1) (y)=f(y) ⇒y=((y^3 +1)/2)  y^3 −2y+1=0 ; (y−1)(y^2 +y−1)=0   { ((y=1)),((y=((−1 ± (√5))/2) )) :}

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

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