Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$
x^3  + 1 = 2\sqrt[3]{{2x - 1}}
$$
I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here?

For $x$ - real
$$2\cdot(2x-1)^\frac13=x^3+1$$
Let $y= (2x-1)^\frac13$
Therefore, $$y^3=2x-1$$
$$x=\frac{(y^3+1)}{2}$$
Then you get
$$2\cdot y=\left(\frac{y^3+1}2\right)^3+1$$
Resolve it for $y$, and then replace find the $x$
Resolving for $y$:


*

*multiply both sides by 8 and you get


$$16\cdot y=(y^3+1)^3+8$$
$$16\cdot y =  y^9+ 3\cdot y^6+3\cdot y^3 +9$$
$$(y-1)(y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 )(y^2 + y - 1)=0$$
Then we get:
$y-1 = 0 \implies y=1$
$y^2 +y - 1=0 \implies y = \frac{1}{2}(-1\pm \sqrt5)$
$y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 =0\implies$  No roots
