Solving $ \left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$ 
Solving :
  $$\left(\frac{  ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$$

My Try :
$$\left(\frac{  ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}=\left(\dfrac{(x+1)^3+2^3}{2^4}\right)^3$$
$$ x^3+a^3=(x+a)^3-3ax(x+a)$$
$$\left(\frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}\right) ^{3}=\left(\dfrac{(x^3+3)(1-6(x^3+1))}{2^4}\right)+1=2x$$
Now what ?

Let $\frac{x^3+1}{2}=y$ and $\sqrt[3]{2x-1}=z.$
Thus, from the given we obtain $\frac{y^3+1}{2}=z.$
Now, let $x>y$.
Thus, $$x>y=\frac{x^3+1}{2}>\frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.
By the same way we'll get a contradiction for $x<y$.
Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$\left\{1,\frac{-1+\sqrt5}{2},\frac{-1-\sqrt5}{2}\right\}$$
