Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically? I'm trying to solve the said equation in the thread title algebraically.
$$(x^3+1)=2\sqrt[3]{2x-1}$$
Cubing both sides and simplifying:
$$x^9+3x^6+3x^3-16x+9 = 0$$
Not sure if this can be solved algebraically?
Edit: WA gives $3$ solutions $x=1,\frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(-1+\sqrt{5})$


$$f(x)=x^9+3x^6+3x^3-16x+9$$

$$f(x)=0\Rightarrow (x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$
where $$x^6+2x^4+2x^3+4x^2+2x+9=x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8>0$$ for $\forall x\in \mathbb{R}$, so we get:
$$(x-1)(x^2+x-1)=0$$
There are three real roots:
$$x_1=1, x_2=\frac{-1-\sqrt{5}}2, x_3=\frac{-1+\sqrt{5}}2$$
