<s>[INST] Roots of the Cubic equation The Cubic formula:
$ax^3+bx^2+cx+d=0$
With under the following conditions:

*

*$a \neq 0 $

*$a,b,c,d \in \Bbb{R} $

We can derive the following formula as a the root of $x$:
$u= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}+\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}  $
$v= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}-\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}  $
$x_1= u + v-{b\over 3a}$


But the proof of derivation of the formula mentioned above was only limited to 1 root. In addition, that formula had no complex conjugates. However as we know, there must be 2 other roots which includes complex conjugates in their formulas.

In reference, Wikipedia: Cubic equation also says that there should be 2 other roots at maximum.
So, altogether the 3 roots are:
$i= \sqrt {-1} $
$ \omega = -{1\over 2} + {{\sqrt 3}i\over 2} $

*

*$x_1= u + v-{b\over 3a}$


*$x_2= {\omega }u + {\omega}^2v -{b\over 3a} $


*$x_3= {\omega }^2 u + {\omega}v -{b\over 3a} $
And unfortunately, I didn't find or know know the proof of any other 2 roots, i.e $x_2$ & $x_3$.
So would you please show me the proof of the other 2 roots of Cubic formula?
Note: Please, no synthetic division. I need proof by formula

 [/INST] Starting with the "depressed" cubic:
$y^3 + py + q = 0$
If you don't have a diminished cubic you can substitute $x = y - \frac {b}{3a}$ which will eliminate the $bx^2$ term from the original cubic.
Next we do a similar substitution:
$y = z-\frac {p}{3z}$
$(z - \frac {p}{3z})^3 + p(z-\frac {p}{3z}) + q = 0\\
z^3 - pz + \frac {p^2}{3z} - (\frac {p}{3z})^3 + pz-\frac {p^2}{3z} + q = 0\\
z^3 - (\frac {p}{3z})^3 + q = 0\\
z^6 + qz^3 - (\frac {p}{3})^3 = 0\\
z^3 = -\frac {q}{2} \pm \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$
We can choose $z^3 = -\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$
with $\frac {1}{z^3} = \frac {-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}{-(\frac {p}{3})^3}$
Or, $-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3} = (-\frac {p}{3z})^3$
Note that $\omega^3 = 1$ has $3$ roots
$\omega_1 = -\frac 12 + \frac {\sqrt 3}{2} i\\
\omega_2 = -\frac 12 - \frac {\sqrt 3}{2} i = \omega_1^2  = \frac {1}{\omega_1}\\
\omega_3 = 1$
$z = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$
with $z$ equal to any of the values of $\omega$ above, and
$y = z - \frac {p}{3z} = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}} + \frac {1}{\omega}{\sqrt[3]{-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$
If $(\frac {q}{2})^2 + (\frac {p}{3})^3 >0 $ there is only one real root.
If $(\frac {q}{2})^2 + (\frac {p}{3})^3 <0 $ the complex terms ultimately cancel out and there are 3 real roots.
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