<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] $m=\sqrt[3]{\frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}}$
Lets choose the positive root for m.
$m=\sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
and we know that this solves:
$q=-(m^3+n^3)$
So lets plug it into $m^3$
$q=\frac{q - \sqrt{q^2-\frac{4p^2}{27}}}{2} - n^3$
$n^3 = \frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}$
Which is the sign flipped other root.
$y = \sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}} + \sqrt[3]{\frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
One more note
$m^3 = \frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}$ has 2 complex roots that should not be forgotten.
$y = \omega \sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}} + \omega\sqrt[3]{\frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
where $\omega$ are the roots of $(z^3-1 = 0)$
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