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!

 A: Write your equation as $$\frac{x^{3}+1}{2}=(2x-1)^{1/3}$$
Now notice that these are inverses of each other. That is, $y=\frac{1}{2}(x^{3}+1)$ is a reflection of $y=(2x-1)^{1/3}$ in the line $y=x$. By symmetry, then, any root must also lie on $y=x$. The easiest way to write this is $$\frac{x^{3}+1}{2}=x \implies x^{3}-2x+1=0$$
Which factorises to $$(x-1)(x^{2}+x-1)=0$$
No heavy machinery required!
A: 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.
A: Consider the function $$f(x)=x^3+1-2{(2x-1)}^{1/3}$$ It is only defined for $x \geq \frac{1}{2}$. If you compute the second derivative $$f''(x)=6 x+\frac{16}{9 (2 x-1)^{5/3}}$$ it is always positive. So, the maximum number of roots is $2$. Since $x=1$ has been identified by inspection, it means that $f(1-\epsilon) \lt 0$ and since $f(\frac{1}{2})=\frac{9}{8}>0$, there is another root located between $\frac{1}{2}$ and $1$.  
As Holographer wrote, the root corresponds to $\frac{-1+\sqrt{5}}{2}$ but I do not know how to find it without knowing the result in advance.
But we can compute it using Newton method. Starting at $x_0=0.6$ and using the classical quadratic scheme $$x_{n+1}=x_n-\frac {f(x_n)}{f'(x_n)}$$ the successive iterates are $0.616459$, $0.618022$, $0.618034$ which is the solution for six significant figures.
