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!


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!
