This question is an exact duplicate of:

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


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 $y=x-1$.


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.


marked as duplicate by Dietrich Burde, Moishe Kohan, Michael Seifert, Community Jan 27 '18 at 14:42

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ I seem to be mising something. You set y=u+v and then derive an expression after cubing. I don't see how you got it. $\endgroup$ – herb steinberg Jan 26 '18 at 17:03
  • $\begingroup$ @herbsteinberg $y^3 = (u+v)^3 = u^3 + v^3 + 3u^2v+3v^2u = u^3 + v^3 + 3uv(u+v) = u^3 + v^3 + 3uvy$ $\endgroup$ – BallBoy Jan 26 '18 at 18:08

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: $$\{2\cos40^{\circ}+1,2\cos80^{\circ}+1,2\cos160^{\circ}+1\}.$$

  • $\begingroup$ Sir I have found a solution using De-Moiveres Law.How to apply them into it? $\endgroup$ – MathsWiz Jan 27 '18 at 14:45
  • $\begingroup$ I like my proof. By the the De-Moiveres we need to get the same result. $\endgroup$ – Michael Rozenberg Jan 27 '18 at 15:43

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 $y$.


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

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

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

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


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