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}$.

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.
