Cube root of $-2+i$

Question

Edit:
My question comes from finding the solutions of this equation using Cardano's method(because our teacher said :D ):
$$x^3-6x+4=0$$
And finally I got:
$$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}⠀+(\sqrt[3]2)\sqrt[3]{-2-\sqrt{-1}}$$
I want to denest and find the cube roots of this two radicals.

According to this links answers:
First link
Second link
Third link
I tried to find this using norm(actually for dinest):
$$\sqrt[3]{-2+i}$$
So the problem's norm is $5$ and I must find a root with norm = $\sqrt[3]{5}$ .

And I found this :
$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)$$
But:
$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)^3\neq {-2+i}$$

Anybody can tell me whats my wrong?
And how to correct that?

Answer 1

Applying Cardano's method to $x^3- 6x+4 =0$ I get:

set $x=u+v$, then

$$x^3 = u^3 + v^3 +3uv(u+v)$$

$$-6x = -6(u+v)$$

so in order to cancel the last terms we set $uv=2$.

So the equation transforms into

$$u^3 + v^3 + 4 = 0; uv=2 $$

Using $v = \frac{2}{u}$ we get

$$u^3 + \frac{8}{u^3} + 4 =0$$

and we set $t = u^3$ and multiply both sides by $t$ to get

$$t^2 + 4t + 8 = 0 \to (t+2)^2 + 4 =0$$

And so $$t+2 = 2i \to t = 2i-2 \text{ or } t+2 = -2i \to t = -2i -2$$

This doesn't seem to lead to your solution?

Added $x=2$ is a solution (the only rational roots are integer divisors of $4$), and the equation factorises to $$(x-2)(x^2 +2x-2)=0$$

Answer 2

$-2+i$ has norm (length) equal to $\sqrt{(-2)^2 + 1^2} = \sqrt{5}=5^{\frac12}$. So the norm of a cube root of it must have norm $(5^{\frac12})^{\frac13} = 5^{\frac16} = \sqrt[6]{5}$.

Now look at the argument of $-2+i$ and take a third of it.

Answer 3

The given number can be presented in the forms of $$-2+i=\sqrt5e^{i\varphi}=\sqrt5\left(\cos\varphi+i\sin\varphi\right),$$ where $$\cos\varphi = -\dfrac2{\sqrt5},\quad \sin\varphi=\dfrac1{\sqrt5},\quad \tan\dfrac\varphi2 = \dfrac{\sin\varphi}{1+\cos\varphi}=\dfrac1{\sqrt5-2}=\sqrt5+2,$$ $$\varphi=2\arctan(\sqrt5+2)\approx2.678.\tag1$$ Taking in account the periodicity $$\sqrt5e^{i\varphi} = \sqrt5e^{i\varphi+2i\pi} = \sqrt5e^{i\varphi-2i\pi},$$ one can get three different values of the required cubic root in the form of $$\sqrt[3]{-2+i}\,=\sqrt[6]5e^{i\large\frac{\varphi+2\pi k}3}, k=0,1,-1.\tag2$$ Then, for example, $$\sqrt[6]5e^{i\large\frac{2\arctan(5+\sqrt2)}3}\approx0.82036+1.01832i.$$

Using of the half-argument in the tangent function allows to avoid problems with the quadrant accounting.