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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?

Arian Ghasemi
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  • Norm is $\sqrt{(-2)^2 + 1^2} = \sqrt{5}.$ – user2661923 Mar 05 '21 at 14:21
  • @user2661923 Added more informatin. – Arian Ghasemi Mar 05 '21 at 15:20
  • @user2661923 I saw that but we ought solve this with Cardano's method only. – Arian Ghasemi Mar 05 '21 at 15:39
  • For what its worth, I deleted my answer in favor of Henno Brandsma's 2nd answer, which I upvoted. See my trailing comment to his answer, which gives $\cos(\pi/12)$ and $\sin(\pi/12)$. When I was verifying part of his work, I used that $$(2 + \sqrt{3})(\sqrt{3} - 1)^2 = (2 + \sqrt{3})(4 - 2\sqrt{3}) = 2 \implies \sqrt{2 + \sqrt{3}}~~(\sqrt{3} - 1) = \sqrt{2}.$$ – user2661923 Mar 05 '21 at 18:27
  • @user2661923 We're not able to know what exactly C numbers mean.We also consider$\sqrt{-1}$ as an abstract thing . In our class examples , the teacher used "guessing" for the cube roots. We cannot use cos or sin. – Arian Ghasemi Mar 06 '21 at 06:30

3 Answers3

2

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 $$(x-2)(x^2 +2x-2)=0$$

Henno Brandsma
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  • @user2661923 We get a third root of $2(i-1)$ and one of $-2(i+1)$ which is close but not equal to what the OP had. Unless I made a calculation mistake. – Henno Brandsma Mar 05 '21 at 17:03
  • @user2661923 These are the $t$ but then use $t= u^3, uv=2$ etc and the original equation has $x=u+v$ as root, not $t$! – Henno Brandsma Mar 05 '21 at 17:07
  • It just hit me; $(x = 2)$ is a root of the equation. – user2661923 Mar 05 '21 at 17:07
  • @user2661923 Indeed a divisor of $4$ as expected. – Henno Brandsma Mar 05 '21 at 17:08
  • @user2661923 One root is $x=\sqrt[3]{2i-2} + \frac{2}{\sqrt[3]{2i-2}}$. Simplify... – Henno Brandsma Mar 05 '21 at 17:11
  • +1 to your answer, which I was finally able to follow. Since the argument of $(1 + i)$ is $(\pi/4)$ and since $\cos(\pi/6) = (\sqrt{3}/2)$ I was able to use (1/2 angle formulas) to compute $$\cos(\pi/12) = (1/2)\sqrt{2 + \sqrt{3}}$$ and $$\sin(\pi/12) = (1/2)\sqrt{2 - \sqrt{3}}.$$ – user2661923 Mar 05 '21 at 18:19
  • For irreducible cubics, https://en.wikipedia.org/wiki/Casus_irreducibilis I guess this example is exempt because it reduces over the rationals... – Will Jagy Mar 05 '21 at 18:19
  • @WillJagy it factors over the reals but not over the rationals. It’s the \Delta>0$ in the casus link. – Henno Brandsma Mar 05 '21 at 18:27
  • I meant your reduction of the original cubic $(x-2)(x^2 +2x-2).$ If we have a cubic that is irreducible over the rationals and has three distinct real roots, we get in a circular mess because we need cube roots of non-real numbers for Cardano, and each such cube root leads to more casus irreducibles. Never ends. – Will Jagy Mar 05 '21 at 18:38
  • Well, examples of that, irreducible but with nice roots (sums of cosines of $2\pi/n$ for $n=7$ or $n=9$) are $x^3 + x^2 - 2x - 1$ for seven, then $x^3 - 3 x + 1$ for nine. These examples show up on this site from time to time. – Will Jagy Mar 05 '21 at 18:54
  • @HennoBrandsma Thanks for the answering effort ,but I was already know that what actually the answers is(e.g 2,...). But the problem is how to denest my cube radicals.This is all I want(even through guessing ). – Arian Ghasemi Mar 06 '21 at 06:37
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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.

Yuri Negometyanov
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$-2+i$ 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.

Henno Brandsma
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