<s>[INST] Simplifying the Solution to the Cubic I am trying to solve the cubic.  I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic.  Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$.  Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\frac{B^2}{3}\right)y+\left(D-\frac{BC}{3}+\frac{2B^3}{27}\right)=0$$
Thus, to make things simpler, i made the substitution $p=C-\frac{B^2}{3}$ and $q=D-\frac{BC}{3}+\frac{2B^3}{27}$ we have the "depressed cubic"
$$y^3+py+q=0$$
Now, using the identity, 
$$(m+n)^3=3mn(m+n)+(m^3+n^3)$$
we let $y=m+n$.  This then translates to $p=-3mn,$ and $q=-(m^3+n^3)$ and gives us a system of equations in $m$ and $n$.  Solving for $n$ gives $n=-\frac{p}{3m}$ and back substituting yields
$$q=-m^3+\frac{p}{3m}\qquad \Rightarrow \qquad m^6+qm^3-\frac{p^3}{27}=0$$ and now we can solve the quadratic for $m$;
$$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}$$
and then that means, by back substitution
$$n=-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
So, I think I am almost here, because now, 
$$y=m+n=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
But how can I simplify this expression?  I know I can back substitute for the original $a,b,c,d$ and solve for $x$.  But this sum looks complicated and my attepts to simplify the sum have not worked.  
 [/INST] For the calculation of the roots of the depressed cubic
$$
y^{\,3}  + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v =  - \frac{p}
{{3\,u}}\quad \omega  = e^{\,i\,\frac{{2\pi }}
{3}} 
$$
where for the radicals you take one value, the real or 
the first complex one (but does not matter which)
then you compute the three solutions as:
$$
y_{\,1}  = u + v\quad y_{\,2}  = \omega \,u + \frac{1}
{\omega }\,v\quad y_{\,3}  = \frac{1}
{\omega }\,u + \omega \,v
$$
Also refer to this post and to this other one.
</s>