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

 A: Use polynomial long division to divide $ax^3+bx^2+cx+d$ by $x-x_1$. This will give you a quadratic, and from there you can find the two remaining roots.
You can do this systematically, using the usual steps for division. If the polynomial is relatively simple, then an alternative would be to write out $$ax^3+bx^2+cx+d = (x-x_1)(ax^2+px+q),$$
then start matching coefficients. For example, looking at $x^2$, we have $b = p-ax_1$, and so on.
A: 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.
A: By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;
and replacing $a, b, c$ by $x, -u, -v$ respectively,
$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.
$u, v$ are known as resolvents.
