Edit:
My question comes from finding the solutions of this equation using Cardano's method(because our teacher said 😀 ):
$$x^3-6x+4=0$$
And finally I got:
$$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}â €+(\sqrt[3]2)\sqrt[3]{-2-\sqrt{-1}}$$
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 :
$$(\sqrt{\sqrt[3]{5}â €-1}â €+i)$$
But:
$$(\sqrt{\sqrt[3]{5}â €-1}â €+i)^3\neq {-2+i}$$
Anybody can tell me whats my wrong?
And how to correct that?
Best Answer
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 $$(x-2)(x^2 +2x-2)=0$$