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!

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 $y^2+y+\frac 1{27}=0$
$u,v= \sqrt[3] {\frac {-1\pm \sqrt{1-\frac 4{27}}}2 }$
$x=u+v=\sqrt[3] {\frac {-1+ \sqrt{\frac {23}{27}}}2 }+\sqrt[3] {\frac {-1- \sqrt{\frac {23}{27}}}2 }$
