Solving Cubic Equation using Cardano's method: $x^3-4x+1=0$

Question

Substituting $x=u+v$, we get $$3uv=4$$ $$u^3+v^3=-1$$ and it follows that, $$u^3=-\frac{1}{2}+\iota \frac{\sqrt{687}}{18}$$ assume, $z=x+\iota y=-\frac{1}{2}+\iota \frac{\sqrt{687}}{18}$ $$r=\sqrt{x^2+y^2}$$ $$r=\frac{8\sqrt{3}}{9}$$ $$\theta=\arctan(-\frac{\sqrt{687}}{9})$$ then the polar form is, $$u^3=r(\cos\theta+i\sin\theta)$$ $$u=r^{1/3}(\cos\frac{\theta+2k\pi}{3}+\iota \sin\frac{\theta+2k\pi}{3}) ,k=0,1,2$$ $$x=u+\frac{4}{3u}$$ $$u^3=r(\cos\theta+i\sin\theta)$$ $$u=r^{1/3}(\cos\frac{\theta+2k\pi}{3}+\iota \sin\frac{\theta+2k\pi}{3}) k=0,1,2 $$ $$x=u+\frac{4}{3u}$$ We can get three solutions of x using k=0,1,2. But this method does not result in correct solutions. Any idea why? Another way I tried, $$v=r^{1/3}(\cos\frac{\theta+2k\pi}{3}-\iota \sin\frac{\theta+2k\pi}{3}) ,k=0,1,2$$ $$x=u+v$$ $$x=2r^{1/3}(\cos\frac{\theta+2k\pi}{3})$$ This method results in correct solutions for x when k=0,1,2 is used, but sign is opposite, which I guess can be fixed when taking signed values of the roots?. Will this method work for all irreducible cases?