Solutions to a Cubic Root I'm looking for a sanity check to my work.
We are supposed to find the solutions of the cubic polynomial, $x^3+3x+5=0$.
Let $x=w-\frac{1}{w}$ and substitute.
Then $(w-\frac{1}{w})^3+3(w-\frac{1}{w})+5=0$.
Then $w^3+5w^3-1=0$.
Finally $w^3=\frac{-5+\sqrt{29}}{2}$ and $\frac{-5-\sqrt{29}}{2}$.
Solution 1: $\sqrt[3]\frac{-5+\sqrt{29}}{2}-\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 2: $\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 3: $\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
 A: Here's a sanity check that shows something is wrong:
$$\sqrt[3]{-5+\sqrt{29}\over2}-\sqrt[3]{-5-\sqrt{29}\over2}=\sqrt[3]{-5+\sqrt{29}\over2}+\sqrt[3]{5+\sqrt{29}\over2}\gt0$$
since $\sqrt{29}\gt5$ (so that the first cube root is positive).  But $x^3+3x+5\gt0$ for all $x\gt0$, so the real roots of $x^3+3x+5$ must be negative.
A: In general Let be $f(x)=x^3+ax+b$
put $x=u-v$, then $f(u-v)=(u^3-v^3)-(3uv-a)(u-v)+b$ and if $f(u,v)=0$, then $(u^3-v^3)+b=0$ and $3uv-a=0$. From second equation we have $v=\frac{a}{3u}$ and then $u^3-(\frac{a}{3u})^3+b=0 \Rightarrow 3^3u^6-a^3+3^3u^3b=0\Rightarrow 3^3y^2+3^3by-a^3=0 ; y=u^3 $, we solve the equation for $y$, then we have $y=\frac{-b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 } \Rightarrow u=\sqrt[3] {y}=\sqrt[3]{\frac{-b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 }} $ and $v^3=u^3+b\Rightarrow v=\sqrt[3]{u^3+b}=\sqrt[3]{\frac{b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 }} $.Therefore first solution of $f(x)$ is $u-v$
and other solutions are complex, i.e. second solution is $\zeta u-\zeta^2v$ and third solution is $\zeta^2 u-\zeta v$. 
In question we have $a=3, b=5$, then $ y=\frac{-5+\sqrt{29}}{2} \Rightarrow u= \sqrt[3] {\frac{-5+\sqrt{29}}{2} }, v= \sqrt[3] {\frac{5+\sqrt{29}}{2} } $ ,so
first solution is $u-v=\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- \sqrt[3] {\frac{5+\sqrt{29}}{2} } $
second solution is $\zeta u-\zeta^2v=(\frac{-1+\sqrt3i}{2})\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- (\frac{-1-\sqrt3i}{2}) \sqrt[3] {\frac{5+\sqrt{29}}{2} }$
third solution is $\zeta^2 u-\zeta v=(\frac{-1-\sqrt3i}{2})\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- (\frac{-1+\sqrt3i}{2}) \sqrt[3] {\frac{5+\sqrt{29}}{2} }$
