Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically?

Question

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

Answer 1

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 $\displaystyle 1, \frac{-1\pm\sqrt5}{2}$.

Answer 2

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

$$f(x)=0\Rightarrow (x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$

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:

$$x_1=1, x_2=\frac{-1-\sqrt{5}}2, x_3=\frac{-1+\sqrt{5}}2$$