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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.

AveryJessup
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    Easy: there is no simpler form. –  Jul 06 '17 at 17:06
  • There's a typo: it's $q^2+\frac{4p^3}{27}$ and not $q^2-\frac{4p^3}{27}$. – José Carlos Santos Jul 06 '17 at 17:07
  • And actually, the solution is more complicated, because $m$ can be a complex number. In this case, you have to resort to trigonometric functions... –  Jul 06 '17 at 17:07
  • @JoséCarlosSantos yes you are right – AveryJessup Jul 06 '17 at 17:08
  • see here https://trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm – Dr. Sonnhard Graubner Jul 06 '17 at 17:10
  • have you tried $$n=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$ ? – poetasis Mar 24 '20 at 18:23

3 Answers3

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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.

José Carlos Santos
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  • Cardano's formula is different though, as it has the sum as cube roots with different signs under the radical and in my derivation, the second term has the cube root int he denominator...so how do you remove the $\pm$ and how do you simplify the denominator? – AveryJessup Jul 06 '17 at 17:30
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    Cardon's formula is usually presented as a sum of two cubic roots $\sqrt[3]A+\sqrt[3]B$, but this is not meant to be taken literally, because then we would get nine roots and not just three. In fact, after we get the first cube root $\sqrt[3]A$, the second one is the cube root $\sqrt[3]B$ of $B$ such that $\sqrt[3]A\times\sqrt[3]B=-\frac p3$. That is basically what you got. And writing $\pm$ instead of simply writing $+$ is just an illusion. You don't get more roots because of that. – José Carlos Santos Jul 06 '17 at 17:38
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$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}}$

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 $(z^3-1 = 0)$

Doug M
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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.

G Cab
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