Cubic Equation With Complex Roots

Question

Please Forgive me for any mistake in the proposal of the problem in advance. Please feel free to edit it.

Problem

Find the roots of the following cubic equation $x^3-3x^2+3=0$.

My Approach

After removing the second degree of $x$ we get $y^3+1-3y=0$ where $y=x-1$.

Then

Taking $y=u+v$ and then cubing both sides I got $y^3-(u^3+v^3)-3uvy=0$. Now after equating the coefficients I got $u^3+v^3=-1 , u^3v^3=1$.

After This I formed the quadratic equation

The quadratic equation formed is $t^2+t+1=0$. The roots are $\dfrac{-1+i\sqrt3}{2}$ and $\dfrac{-1-i\sqrt3}{2}$.

After this I take

$r\cos\theta=\dfrac{-1}{2}$ and $r\sin\theta=\dfrac{\sqrt3}{2}$ I got $\theta=60^\circ$.

However I don't know how to approach further.Please tell me if I am on the right track or not.

Note:-I use Cardano's Method.

Any help is welcome.

Note:-This Question is different from the duplicate as this problem states the problem after we form the quadratic equation and try to get the cube roots of a complex number.

Answer 1

Let $x=2\cos\alpha+1$.

Thus, $$8\cos^3\alpha+12\cos^2\alpha+6\cos\alpha+1-12\cos^2\alpha-12\cos\alpha-3+3=0$$ or $$4\cos^3\alpha-3\cos\alpha=-\frac{1}{2}$$ or $$3\alpha=\pm120^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$ or $$\alpha=\pm40^{\circ}+120^{\circ}k,$$ which gives the answer: $$\{2\cos40^{\circ}+1,2\cos80^{\circ}+1,2\cos160^{\circ}+1\}.$$

Answer 2

You have found $t$. But what is $t$? If you look back at the way you derived the quadratic equation, $t = u^3$ (or $v^3$). So, to find $u$, you need to take the cube root of $t$.

To take the cube root of a complex number expressed in terms of $r$ and $\theta$, you take the regular cube root of $r$, and take $\frac{\theta}{3}$ plus multiples of $\frac{2\pi}{3}$ giving you three cube roots. This gives possible values of $u$ and $v$. Then you add them to get $y$.

Answer 3

Using Nickalls' "A new approach to solving the cubic: Cardan's solution revealed" :

$$x^3-3x^2+3 = 0$$

has its N-point at $x_N = \dfrac{-b}{3a} = 1$, so the substitution $x = z + x_N = z+1$ can be used to depress the cubic, resulting in:

$$ z^3 - 3z + 1 = az^3 - 3a\delta^2z + y_N = 0$$

From this we see Nickalls' parameters $y_N = 1$ and $\delta^2 = 1$.

Thus Nickalls' parameter $h = 2a\delta^3 = 2$.

Since $y_N^2 < h^2$, you have the case of three, distinct, real roots. To avoid having to find the cube root of complex numbers, the method for this case uses a trigonometric substitution (as @Michael_Rozenberg shows in his answer).

Proceeding with the formulae Nickalls has derived:

$$\cos 3\theta = \dfrac{-y_N}{h} = -\dfrac{1}{2}$$ we have $$3\theta = \dfrac{2\pi}{3}$$ or $$\theta = \dfrac{2\pi}{9}$$

and the three roots of the original cubic polynomial are

$\alpha = x_N + 2\delta\cos(\theta) = 1 + 2\cos\left(\dfrac{2\pi}{9}\right)$

$\beta = x_N + 2\delta\cos\left(\theta + \dfrac{2\pi}{3}\right) = 1 + 2\cos\left(\dfrac{8\pi}{9}\right)$

$\gamma = x_N + 2\delta\cos\left(\theta + \dfrac{4\pi}{3}\right) = 1 + 2\cos\left(\dfrac{14\pi}{9}\right)$