<s>[INST] How to solve the following? $ x^3+1=2{(2x-1)}^{1/3} $. Find all the real solutions of
$$x^3+1=2{(2x-1)}^{1/3}  $$  

I tried to cube both sides but got messed up with a nine degree equation! Please help.
  Thanks in advance!

 [/INST] We are given 
$$x^3+1=2{(2x-1)}^\frac{1}{3}$$   
Substitute $(2x-1)=y$  
$\implies x^3=2y-1$  
Also,  
$y^3=2x-1$ [Using the initial substitution]  
Subtracting the above two equations, we get,  
$x^3-y^3=-2(x-y)$  
$\implies x=y$  
or,  
$x^2+xy+y^2=-2$   
But since $x^2+y^2 \ge xy$ and L.H.S. is thus positive,   
$\implies x^2+y^2+xy=-2$ yields no solution.   
Now, as $x=y$,  
$\implies x=(2x-1)^{\frac{1}{3}}$  
$\implies x^3-2x+1=0$  
Now, using rational root theorem, $x=1$ satisfies the above equation. So, by dividing with $(x-1)$, we obtain  
$(x-1)(x^2+x-1)=0$  
$\implies x=1,\frac{-1 \pm \sqrt{5}}{2}$ are the required solutions.
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