Solve $x^3 - x + 1 = 0$ Solve $x^3 - x + 1 = 0$, this cannot be done through elementary methods.
Although, this is way out of my capabilities, I would love to see a solution (closed form only). 
Thanks!
 A: Let $x=u+v$ and note that $(u+v)^3=3uv(u+v)+(u^3+v^3)$
This has the required form if $3uv=1$ and $u^3+v^3=-1$
So $u^3v^3=\frac 1{27}$, and $u^3, v^3$ are roots of $y^2+y+\frac 1{27}=0$
$u,v= \sqrt[3] {\frac {-1\pm \sqrt{1-\frac 4{27}}}2 }$
$x=u+v=\sqrt[3] {\frac {-1+ \sqrt{\frac {23}{27}}}2 }+\sqrt[3] {\frac {-1- \sqrt{\frac {23}{27}}}2 }$
A: Let $$x=\frac{a}{y}+y$$ then $$y=\frac12 \left (x+\sqrt{x^2-4a} \right )$$
Then you have $$1-y-\frac{a}{y}+\left (y+\frac{a}{y}\right )^3 =0\stackrel{\cdot y^3}{\iff} y^6+(3a-1)y^4+y^3+(3a^2-a)y^2+a^3=0\;\stackrel{a=\frac13,z=y^3}{\iff}\; z^2+z+\frac{1}{27}=0$$
Solve and rewind all the substitutions(not the $a$ one, I kept it along to see what would eventually fit). Didn't pay attention to any restrictions so you'll have to check each value too. I would like to know where my method is non-elementary too.
PS: The $a$ part is to show why this Vieta's substitution variant works.
A: There is a standard way of solving these kind of equations: with the Cardano method. Look here
