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 $x=1,\frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(-1+\sqrt{5})$

Rearrange as $$\underbrace{\frac{x^3+1}{2}}_{f(x)}=\underbrace{\sqrt[3]{2x-1}}_{g(x)}$$
Since $f(x)$ is a bijective function on $\mathbb R$, it must have an inverse. But note that the inverse of $f(x)$ is $g(x)$. Thus, if the two curves intersect, they must intersect ON the line $y=x$ (because $f$ and $g$ are mirror images about this line).
Thus we solve the two curves $y=\dfrac{1+x^3}{2}$ and $y=x$ to get $$x^3-2x+1=0$$ which boils down to $$(x-1)(x^2+x-1)=0$$ which can be readily solved to get solutions $\displaystyle 1, \frac{-1\pm\sqrt5}{2}$.
