How can I find the roots of $x^3-3x+1$ using Cardano's formula?
So far, I found that
$$x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}$$ $$x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}$$ $$x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt[3]{\frac{-1}{2}-\frac{i\sqrt{-3}}{2}}$$
I am now trying to express each cubic radicand in their exponential form.
Euler's formula : $e^{i\theta} = \cos \theta + i\sin \theta$.
I have $r = |z| = 1$.
However, when I try to find the angle, I get
$$\theta = \arccos \frac{-1}{2} = 2\pi/3$$
but at the same time
$$\theta = \arcsin \frac{\sqrt{-3}}{2} = \pi/3$$
Shouldn't the two angles be the same for one radicand?
Plus, once I will have expressed the two radicands in their exponential form, what is the next step?
$$x = \sqrt[3]{e^{i2\pi/3}} + \sqrt[3]{e^{i4\pi/3}}$$ $$x = e^{i2\pi/9} + e^{i4\pi/9}$$
Am I on the right track?
Use the Cardano approach:
Let $$ x=u+v $$ then we will get
$$u^3+v^3+3(uv-1)(u+v)+1=0$$
by setting $uv-1=0$ we get $$uv=1$$ and $$u^3+v^3=-1$$ by cubing the first equation we obtain a simple quadratic system in terms of $u^3$ and $v^3$ which you can be reduced to a quadratic equation in terms of $u^3$ for example: $$u^3+\frac{1}{u^3}+1=0$$
Then use the Quadratic formula to find $u^3$ and $v^3=-1-u^3$
After extracting cubic roots of the obtained solution you need to take into account that $uv$ is real. This will give the appropriate pairs of $u$ and $v$ whose sums will yield all three roots of the equation.