<s>[INST] 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?
 [/INST] We have $$x^3+1=2(2x-1)^{1/3}\iff x^3=2(2x-1)^{1/3} -1.$$
Here, setting $y=(2x-1)^{1/3}$ gives us
$$y^3=2x-1 \ \ \text{and}\ \ x^3=2y-1.$$
Hence, we have
$$\begin{align}y^3-x^3=(2x-1)-(2y-1)&\Rightarrow (y-x)(y^2+yx+x^2)=2(x-y)\\&\Rightarrow  (y-x)(y^2+yx+x^2+2)=0\\&\Rightarrow  (y-x)\{(x+(y/2))^2 + (3/4)y^2+2\}=0\\&\Rightarrow  y=x.\end{align}$$
Hence, we have
$$x^3=2x-1\iff (x-1)(x^2+x-1)=0\iff x=1,\frac{-1\pm\sqrt 5}{2}.$$
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