Solving third-degree equations (cubic equations)

Existence of obvious root(s)

If there are one or more obvious roots, then the polynomial can be factored and reduced to a lower degree.

Determination of the number of real roots according to the cases

In the general case, we can use a graphical approach to determine the number of roots according to the different cases.

We start from the equation \( (1) \) :

We factor by \( a \) :

From this point on, we will keep in mind that the general form \((2)\) contains \(a\) as a factor, but we will study the function inside the parenthesis:

We then seek to eliminate the second degree term.

The expression \((2^*)\) then becomes:

We will then study the function \((f)\):

Let us note from now on that \(f\) is an odd function , being the sum of odd functions.

Next, let's try to derive this function:

Let's factor this derivative \(f'\).

And let us seek these roots:

Let there be two roots:

So, we can rewrite it in its factorized form:

$$ \chi $$	$$ -\infty $$		$$ \chi_1 = -\sqrt{- \frac{p}{3}} $$		$$ \chi_2 = \sqrt{- \frac{p}{3}} $$		$$ +\infty $$
$$ \chi + \sqrt{- \frac{p}{3}} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ 0 $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$
$$ \chi - \sqrt{- \frac{p}{3}} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ 0 $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$
$$ f'(\chi) = \left(\chi + \sqrt{- \frac{p}{3}} \right)\left(\chi -\sqrt{- \frac{p}{3}} \right) $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ 0 $$	$$ \textcolor{rgb(232 124 124)}{-} $$	$$ 0 $$	$$ \textcolor{rgb(93 183 129)}{+} $$	$$ \textcolor{rgb(93 183 129)}{+} $$
$$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$		$$ f(\chi_1) $$		$$ f(\chi_2) $$		$$ + \infty $$

As the function \(f\) change de sens de variations changes direction of variations before \(\chi_1\) and after \(\chi_2\), it therefore has two inflection point in \(\chi_1\) and \(\chi_2\).

And as we know that \( P_3(\chi)\) is worth:

The variation table of \(P_3(\chi)\) will be that of \(f(\chi)\) if \((a > 0)\), but will be reversed if \((a < 0)\).

Let's look for the values of the images \( f\left(\chi_1 \right) \) and \( f\left(\chi_2 \right) \) :

To know the number of roots of \(f(\chi)\), we try to know the sign at a given instant of the product of these two images.

which we will call third-degree discriminant \(\Delta_3\):

We will then have three sub-cases to consider:

• \( \alpha) \) if \(\Delta_3 > 0 \)

  If the discriminant is positive, this means that \(f\left(\chi_1\right)\) and \(f\left(\chi_2\right)\) are of the same sign, therefore:

  • etiher both positive

    According to the variation table of \(f\), there will be only one real root, located between \((\chi = -\infty)\) and \((\chi = \chi_1)\).

    $$ \chi $$	$$ -\infty $$	$$ \textcolor{#6187B2}{\alpha_1} $$	$$ \chi_1 = -\sqrt{- \frac{p}{3}} $$		$$ \chi_2 = \sqrt{- \frac{p}{3}} $$		$$ +\infty $$
    $$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$	$$ \textcolor{#6187B2}{f(\alpha_1) = 0} $$	$$ \textcolor{rgb(93 183 129)}{f(\chi_1) > 0} $$		$$ \textcolor{rgb(93 183 129)}{f(\chi_2) > 0} $$		$$ + \infty $$

    
  • or both negative

    According to the variation table of \(f\), there will be only one real root, located between \((\chi = \chi_2)\) and \((\chi = +\infty)\).

    $$ \chi $$	$$ -\infty $$		$$ \chi_1 = -\sqrt{- \frac{p}{3}} $$		$$ \chi_2 = \sqrt{- \frac{p}{3}} $$	$$ \textcolor{#6187B2}{\alpha_1} $$	$$ +\infty $$
    $$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$		$$ \textcolor{rgb(232 124 124)}{f(\chi_1) < 0} $$		$$ \textcolor{rgb(232 124 124)}{f(\chi_2) < 0} $$	$$ \textcolor{#6187B2}{f(\alpha_1) = 0} $$	$$ + \infty $$

  In both cases, there will be only one real root .

• \( \beta) \) if \( \Delta_3 = 0 \)

  If the discriminant is zero, this means that one of the two images is zero, because according to the variation table, we always have:

  $$ f(\chi_1) > f(\chi_2) $$
  • \( f(\chi_1) = 0 \)

    In this case, there will be a double root for \((\chi = \chi_1)\), and another root for \((\chi_2 < \chi < +\infty)\).

    $$ \chi $$	$$ -\infty $$		$$ \textcolor{#6187B2}{\alpha_1 / \alpha_2 = \chi_1} $$		$$ \chi_2 = \sqrt{- \frac{p}{3}} $$	$$ \textcolor{#6187B2}{\alpha_3} $$	$$ +\infty $$
    $$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$		$$ \textcolor{#6187B2}{f(\chi_1) = 0} $$		$$ \textcolor{rgb(232 124 124)}{f(\chi_2) < 0} $$	$$ \textcolor{#6187B2}{f(\chi_2) = 0} $$	$$ + \infty $$

    
  • \( f(\chi_2) = 0 \)

    Dans ce cas symétrique, il y aura une racine simple pour \((\chi_2 < \chi < +\infty)\), ainsi qu'une racine double pour \((\chi = \chi_1)\).

    $$ \chi $$	$$ -\infty $$	$$ \textcolor{#6187B2}{\alpha_1} $$	$$ \chi_1 = -\sqrt{- \frac{p}{3}} $$		$$ \textcolor{#6187B2}{\alpha_2 / \alpha_3 = \chi_2} $$		$$ +\infty $$
    $$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$	$$ \textcolor{#6187B2}{f(\alpha_1) = 0} $$	$$ \textcolor{rgb(93 183 129)}{f(\chi_1) > 0} $$		$$ \textcolor{#6187B2}{f(\chi_2) = 0} $$		$$ + \infty $$

    

  In both cases, we will have a double real root and a simple real root .

  The double root necessarily lies on the plateau, because in the other case the function is strictly increasing .

• \( \gamma) \) if \( \Delta_3 < 0 \)

  If the discriminant is negative, the two images \(f(\chi_1)\) et \(f(\chi_2)\) are of opposite signs.

  As we know that:

  $$ f(\chi_1) > f(\chi_2) $$

  there is only one case, and we will distinguish three real roots .

  $$ \chi $$	$$ -\infty $$	$$ \textcolor{#6187B2}{\alpha_1} $$	$$ \chi_1 = -\sqrt{- \frac{p}{3}} $$	$$ \textcolor{#6187B2}{\alpha_2} $$	$$ \chi_2 = \sqrt{- \frac{p}{3}} $$	$$ \textcolor{#6187B2}{\alpha_3} $$	$$ +\infty $$
  $$ f(\chi) = \chi^3 + p\chi + q $$	$$ - \infty $$	$$ \textcolor{#6187B2}{f(\alpha_1) = 0} $$	$$ \textcolor{rgb(93 183 129)}{f(\chi_1) > 0} $$	$$ \textcolor{#6187B2}{f(\alpha_2) = 0} $$	$$ \textcolor{rgb(232 124 124)}{f(\chi_2) < 0} $$	$$ \textcolor{#6187B2}{f(\alpha_3) = 0} $$	$$ + \infty $$

  
• \( \delta) \) summary

  In summary, here are the cases according to the result of the discriminant \(( \Delta_3 = q^2 + \frac{4p^3}{27} )\):

  Sign of the discriminant \( \Delta_3 \)	Numbers of roots
  $$ \Delta_3 > 0 $$	1 real root
  $$ \Delta_3 = 0 $$	3 real roots (including one double)
  $$ \Delta_3 < 0 $$	3 real roots

Resolution

Let's briefly summarize the path taken in the previous point. We started from the equation to be solved \((1)\):

Then, we factorized by \(a\) :

The \(a\) factor not changing the roots, we can get rid of it:

and make a first change of variable.

We then arrived at:

Which gives us:

We develop the expression:

Now multiplying by \(w^3\), we obtain that:

We now have a second-degree equation by once again introducing a new variable:

So, the expression\((4^*)\) now becomes:

With this second-degree equation , we can calculate the second degree discriminant, which we will call as above \(\Delta_3\) :

1. if \( (\Delta_3 > 0) \)

   With the equation \((6)\):

   $$ z^2 + qz - \frac{p^3}{27} = 0 \qquad (6) $$

   If \( (\Delta_3 > 0) \), we obtain two real solutions:

   $$ z_1 / z_2 = \frac{- q \pm \ \sqrt{\Delta_3}}{2} $$
   $$ \Bigl(\text{with } \ \Delta_3 = q^2 + \frac{4p^3}{27} \Bigr) $$

   Which can be arranged in:

   $$ z_1 / z_2 = -\frac{q}{2} \pm \frac{ \sqrt{q^2 + \frac{4p^3}{27}}}{2} $$
   $$ z_1 / z_2 = -\frac{q}{2} \pm \frac{ \sqrt{q^2 + \frac{4p^3}{27}}}{\sqrt{4}} $$
   $$ z_1 / z_2 = -\frac{q}{2} \pm \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } $$
   $$ z_1 / z_2 = -\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 } $$

   We made these different variable changes as a result:

   $$ \left[ X = \chi - \frac{b}{3a} \right] \ puis \ \biggl[ \chi = w - \frac{p}{3w} \biggr] \ puis \ \Bigl[ w^3 = z \Bigr] $$

   You must therefore reassemble them in the opposite direction:

   $$ \Bigl[ z \longmapsto w \Bigr] \ puis \ \Bigl[ w \longmapsto \chi \Bigr] \ puis \ \Bigl[ \chi \longmapsto X \Bigr] $$
   1. \(z \longmapsto w \)

      Taking the cube root of \(z\), we will have as coefficient the three roots of unity which are three unit roots of a complex number which are \( \Bigl \{1, \ e^{\frac{2i\pi}{3}}, \ e^{-\frac{2i\pi}{3}} \Bigr \} \), which gives us six roots in all:

      $$ w_1 / w_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      $$ w_3 / w_4 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      $$ w_5 / w_6 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$
   2. \(w \longmapsto \chi \)

      The six expressions then become:

      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p}{3\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }}} $$

      $$ \chi_3 / \chi_4 = e^{\frac{2i\pi}{3}} \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p}{3 e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }}} $$

      $$ \chi_5 / \chi_6 = e^{-\frac{2i\pi}{3}} \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p}{3 e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }}} $$

      Then, we multiply the numerators and denominators by their respective conjugates.

      For the first pair of solutions \(\chi_1 / \chi_2 \) :

      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p \textcolor{rgb(232 124 124)}{\sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }}} }{3\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} \textcolor{rgb(192 52 52)}{\sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }}} } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} }{3\sqrt[3]{\left( -\frac{q}{2} \right)^2 - \left(\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right) } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} }{3\sqrt[3]{\underbrace{\left(\frac{q}{2} \right)^2 - \left(\frac{q}{2}\right)^2 } _\text{= 0} - \left( \frac{p}{3} \right)^3 } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} }{3\sqrt[3]{ \left(- \frac{p}{3} \right)^3 } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} - \frac{p \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} }{-\frac{3p}{3}} $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      These two roots are actually one, because of double signs \(\pm\) and \(\mp \) present in each of the first two terms:

      $$ \chi_1 = \chi_2 = \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      For the other two pairs of solutions \((\chi_3 / \chi_4, \ \chi_5 / \chi_6)\) , we follow the same approach, and after calculations we arrive at the following forms:

      $$ \chi_3 / \chi_4 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      $$ \chi_5 / \chi_6 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      For them too, we can group them together because they are two by two identical:

      $$ X_3 = X_6 = e^{\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{-\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$

      $$ X_4 = X_5 = e^{\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{-\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} $$
   3. \(\chi \longmapsto X \)

      For this last step, we just have to add to each solution the term \(-\frac{b}{3a}\).

      $$ X_1 = X_2 = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + \sqrt[3]{-\frac{q}{2} \mp \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} -\frac{b}{3a} $$
      $$ X_3 = X_6 = e^{\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{-\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} -\frac{b}{3a} $$
      $$ X_4 = X_5 = e^{\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} + e^{-\frac{2i\pi}{3}} \ \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 }} -\frac{b}{3a} $$
   4. conclusion

      We saw this equality above:

      $$ \frac{- q \pm \ \sqrt{\Delta_3}}{2} = -\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 } $$

      So, we put this equality back in place for the presentation of the solutions.

      In this case of \( (\Delta_3 > 0) \) we then have that:

      • a simple real root
        $$ X_1 = \sqrt[3]{\frac{-q + \sqrt{\Delta_3}}{2}} + \sqrt[3]{\frac{-q - \sqrt{\Delta_3}}{2}} - \frac{b}{3a} $$
      • two complex roots
        $$ C_2 = e^{\frac{2i\pi}{3}} \ \ \sqrt[3]{\frac{-q - \sqrt{\Delta_3}}{2}} + e^{-\frac{2i\pi}{3}} \ \ \sqrt[3]{\frac{-q + \sqrt{\Delta_3}}{2}} - \frac{b}{3a} $$

        $$ C_3 = e^{\frac{2i\pi}{3}} \ \ \sqrt[3]{\frac{-q + \sqrt{\Delta_3}}{2}} + e^{-\frac{2i\pi}{3}} \ \ \sqrt[3]{\frac{-q - \sqrt{\Delta_3}}{2}} - \frac{b}{3a} $$
      $$ \text{with } \ \left(\Delta_3 = q^2 + \frac{4p^3}{27} \right) $$
      $$ et \ \left \{ \begin{gather*} p = \frac{-b^2 + 3ac}{3a^2} \\ \\ q = \frac{2b^3 + 27a^2d - 9abc}{27a^3} \end{gather*} \right \} $$

      Hence, the polynomial \(P_3(X)\) admits the following factorization:

      $$ P_3(X) = a(X - X_1)(X - C_2)(X - C_3) $$
2. if \( (\Delta_3 = 0) \)

   With the equation \((6)\):

   $$ z^2 + qz - \frac{p^3}{27} = 0 \qquad (6) $$

   if \( (\Delta_3 = 0) \), we have a real double solution:

   $$ z_0 = -\frac{q}{2} $$

   In the same way as above, we go back through the different variable changes one after the other.

   1. \(z \longmapsto w \)
      $$ w_1 = \sqrt[3]{-\frac{q}{2}} $$

      $$ w_2 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} $$

      $$ w_3 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} $$
   2. \(w \longmapsto \chi \)

      For the fisrt solution \(\chi_1 \) :

      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} - \frac{p}{3\sqrt[3]{-\frac{q}{2}}}$$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} - \frac{p}{3\sqrt[3]{-\frac{q}{2}}} $$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} - \frac{p}{3}\frac{\textcolor{rgb(232 124 124)}{\sqrt[3]{\left(-\frac{q}{2} \right)^2}}}{\sqrt[3]{-\frac{q}{2}} \textcolor{rgb(232 124 124)}{\sqrt[3]{\left(-\frac{q}{2} \right)^2}}} $$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} - \frac{p}{3}\frac{\sqrt[3]{\left(-\frac{q}{2} \right)^2}}{-\frac{q}{2}} $$

      But as:

      $$ \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 } = \frac{\Delta_3}{2} $$

      (voir plus haut)

      And that we have as hypothesis that \( \Delta_3 = 0 \), then:

      $$\Delta_3 = 0 \Longrightarrow 2 \sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 } = 0 $$

      And so that:

      $$\sqrt{\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 } = 0 $$

      Which also implies that:

      $$\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 = 0 $$

      $$\left(\frac{q}{2}\right)^2 = - \left( \frac{p}{3} \right)^3 $$

      $$\sqrt[3]{ \left(\frac{q}{2}\right)^2 } = \sqrt[3]{ - \left( \frac{p}{3} \right)^3 } $$

      $$\sqrt[3]{ \left(\frac{q}{2}\right)^2 } = -\frac{p}{3} $$

      We can then replace in the previous expression:

      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} - \frac{p}{3}\frac{\sqrt[3]{\left(-\frac{q}{2} \right)^2}}{-\frac{q}{2}} $$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} + \frac{ \sqrt[3]{ \left(\frac{q}{2}\right)^2 } \sqrt[3]{\left(-\frac{q}{2} \right)^2}}{-\frac{q}{2}} $$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} + \frac{ \sqrt[3]{ \left(-\frac{q}{2}\right)^2 } \sqrt[3]{\left(-\frac{q}{2} \right)^2}}{-\frac{q}{2}} $$

      Now, we can make the power formulas work:

      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} + \left(-\frac{q}{2}\right)^{\frac{4}{3}} \left(-\frac{q}{2}\right)^{-1 } $$
      $$ \chi_1 = \sqrt[3]{-\frac{q}{2}} + \left(-\frac{q}{2}\right)^{\frac{1}{3}} $$
      $$ \chi_1 = 2\sqrt[3]{-\frac{q}{2}} $$

      For the other two solutions :

      $$ w_2 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} \chi_2 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} $$

      $$ w_3 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} \chi_3 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} + e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} $$

      These two solutions are one and the same solution:

      $$ \chi_2 = \chi_3 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2}} $$
      $$ \chi_2 = \chi_3 = \sqrt[3]{-\frac{q}{2}} \left(e^{\frac{2i\pi}{3}} + e^{-\frac{2i\pi}{3}} \right) $$
      $$ \chi_2 = \chi_3 = -\sqrt[3]{-\frac{q}{2}} $$
   3. \(\chi \longmapsto X \)
      $$ X_1 = 2\sqrt[3]{-\frac{q}{2}} - \frac{b}{3a} $$
      $$ X_2 = X_3 = -\sqrt[3]{-\frac{q}{2}} - \frac{b}{3a} $$
   4. conclusion

      In this case of \( (\Delta_3 = 0) \) we then have that:

      • a simple real root
        $$ X_1 = 2\sqrt[3]{-\frac{q}{2}} - \frac{b}{3a} $$
      • a double real root
        $$ X_2 = X_3 = -\sqrt[3]{-\frac{q}{2}} - \frac{b}{3a} $$
      $$ \text{with } \ \left(q = \frac{2b^3 + 27a^2d - 9abc}{27a^3} \right) $$

      Hence, the polynomial \(P_3(X)\) admits the following factorization:

      $$ P_3(X) = a(X - X_1)(X - X_2)^2 $$
      
      

      Furthermore, starting from these two expressions \((3)\) and its derivative \((4)\):

      $$ \chi^3 + p\chi + q = 0 \qquad (3)$$
      $$ 3\chi^2 + p = 0 \qquad (4) $$
      $$ p = -3\chi^2 \qquad (4^*) $$

      Now, by substituting \((4^*)\) into \((3)\) :

      $$ \chi^3 + -3\chi^2\chi + q = 0 $$
      $$ \chi^3 + -3\chi^3 + q = 0 $$
      $$ -2\chi^3 + q = 0 $$
      $$ \chi^3 = \frac{q}{2} \qquad (\chi^3) $$

      And with the previous relationship \((4^*)\), we do also have this:

      $$ \chi^2 = -\frac{p}{3} \qquad (\chi^2) $$

      Then by multiplying by \(\chi\) on both sides:

      $$ \chi^3 = -\frac{p}{3}\chi \qquad (\chi^3)^* $$

      Finally, with both equal expressions \((\chi^3)\) and \((\chi^3)^*\), we can deduce a relationship between \(p\) and \(q\) :

      $$ \frac{q}{2} = -\frac{p}{3}\chi $$
      $$ \chi = -\frac{3q}{2p} \qquad \Bigl(\chi = f(p, q)\Bigr) $$

      But, with the expression \((\chi^3)\), we saw that:

      $$ -\chi^3 = -\frac{q}{2} \qquad (\chi^3) $$

      So, by taking the third root, we have now:

      $$ \sqrt[3]{-\chi^3} = \sqrt[3]{-\frac{q}{2}} $$

      $$ -\chi = \sqrt[3]{-\frac{q}{2}} $$

      So, by replacing with the value found in \(\Bigl(\chi = f(p, q)\Bigr)\), we finally determined that:

      $$ \frac{3q}{2p} = \sqrt[3]{-\frac{q}{2}} $$

      We can now replace the value of \(\sqrt[3]{-\frac{q}{2}}\) by \(-\frac{3q}{2p}\) in the three relationships \(X_1\), \(X_2\) and \(X_3\).

      So, these three solutions can also be written as:

      • a simple real root
        $$ X_1 = \frac{3q}{p} - \frac{b}{3a} $$
      • a double real root
        $$ X_2 = X_3 = \frac{3q}{2p} - \frac{b}{3a} $$
3. if \( (\Delta_3 < 0) \)

   With the equation \((6)\):

   $$ z^2 + qz - \frac{p^3}{27} = 0 \qquad (6) $$

   If \( (\Delta_3 < 0) \), we do obtain two complex solutions:

   $$ z_1 = \frac{- q \pm i\sqrt{|\Delta_3|}}{2} $$

   Which can be arranged as previously in:

   $$ z_1 / z_2 = -\frac{q}{2} \pm i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| } $$

   In the same way as above, we go back through the different variable changes one after the other.

   1. \(z \longmapsto w \)
      $$ w_1 / w_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$

      $$ w_3 / w_4 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$

      $$ w_5 / w_6 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$
   2. \(w \longmapsto \chi \)

      The six expressions then become:

      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p}{3\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }}} $$

      $$ \chi_3 / \chi_4 = e^{\frac{2i\pi}{3}} \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p}{3 e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }}} $$

      $$ \chi_5 / \chi_6 = e^{-\frac{2i\pi}{3}} \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p}{3 e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }}} $$

      For the first couple of solutions \(\chi_1 / \chi_2 \) :

      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \textcolor{rgb(232 124 124)}{\sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }}} }{3\sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} \textcolor{rgb(232 124 124)}{\sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }}} } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} }{3\sqrt[3]{\left( -\frac{q}{2} \right)^2 + \left|\left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} }{3\sqrt[3]{\left( -\frac{q}{2} \right)^2 + -\left(\frac{q}{2}\right)^2 - \left( \frac{p}{3} \right)^3 } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} }{3\sqrt[3]{\underbrace{\left(\frac{q}{2} \right)^2 + -\left(\frac{q}{2}\right)^2} _\text{ = 0 } - \left( \frac{p}{3} \right)^3 } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} }{3\sqrt[3]{- \left( \frac{p}{3} \right)^3 } } $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{p \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} }{-\frac{3p}{3}} $$
      $$ \chi_1 / \chi_2 = \sqrt[3]{-\frac{q}{2} \ \pm \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + \sqrt[3]{-\frac{q}{2} \ \mp\ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$

      These two solutions are one and the same solution.

      $$ \chi_1 = \chi_2 = \sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + \sqrt[3]{-\frac{q}{2} \ - \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$

      For the other two pairs of solutions \((\chi_3 / \chi_4, \ \chi_5 / \chi_6)\) , we follow the same approach, and after calculations we arrive at the following forms:

      $$ \chi_3 / \chi_4 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ - \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$

      $$ \chi_5 / \chi_6 = e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ - \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} $$
   3. \(\chi \longmapsto X \)
      $$ X_1 = X_2 = \sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + \sqrt[3]{-\frac{q}{2} \ - i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{b}{3a} $$
      $$ \chi_3 = \chi_4 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ - \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{b}{3a} $$

      $$ \chi_5 = \chi_6 = e^{\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ - \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} + e^{-\frac{2i\pi}{3}}\sqrt[3]{-\frac{q}{2} \ + \ i\sqrt{\left| \left(\frac{q}{2}\right)^2 + \left( \frac{p}{3} \right)^3 \right| }} - \frac{b}{3a} $$
   4. conclusion

      In the same way as above, we get rid of the absolute value in the presentation of the results.

      In this case of \( (\Delta_3 < 0) \) we then have that:

      • three real roots
        $$ X_1 = \sqrt[3]{\frac{-q \ + \ i\sqrt{-\Delta_3}}{2}} + \sqrt[3]{\frac{-q \ - \ i\sqrt{-\Delta_3}}{2}} - \frac{b}{3a} $$

        $$ X_2 = e^{\frac{2i\pi}{3}}\sqrt[3]{\frac{-q \ + \ i\sqrt{-\Delta_3}}{2}} + e^{-\frac{2i\pi}{3}}\sqrt[3]{\frac{-q \ - \ i\sqrt{-\Delta_3}}{2}} - \frac{b}{3a} $$

        $$ X_3 = e^{\frac{2i\pi}{3}}\sqrt[3]{\frac{-q \ - \ i\sqrt{-\Delta_3}}{2}} + e^{-\frac{2i\pi}{3}}\sqrt[3]{\frac{-q \ + \ i\sqrt{-\Delta_3}}{2}} - \frac{b}{3a} $$
      $$ \text{with } \ \left(\Delta_3 = q^2 + \frac{4p^3}{27} \right) $$
      $$ et \ \left \{ \begin{gather*} p = \frac{-b^2 + 3ac}{3a^2} \\ \\ q = \frac{2b^3 + 27a^2d - 9abc}{27a^3} \end{gather*} \right \} $$

      Even if they appear in complex form , after solving these solutions definitely become real .

      Hence, the polynomial \(P_3(X)\) admits the following factorization:

      $$ P_3(X) = a(X - X_1)(X - X_2)(X - X_3) $$