Simplifying the Solution to the Cubic

Question

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.

Answer 1

You can't. This is as simple as you can get, unless you want to re-obtain Cardano's formula, which is basically what you got.

Answer 2

$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}}$

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 $(z^3-1 = 0)$

Answer 3

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.