Some time ago, I got a question of the form $\sqrt{a+(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} }$, which after cubing I realized that $(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} = -a $. That set my instincts, and I figured out that that all cubic of the form $x^3 = a + bx$ must have a solution of the form $(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3}$, so I came to the formula:

$$(\frac{a}{2} - \sqrt{\frac{27a^2-4b^3}{108}})^\frac{1}{3} + (\frac{a}{2} + \sqrt{\frac{27a^2-4b^3}{108}})^\frac{1}{3}$$

I have still not managed to verify the formula, especially because of my faulty labelling of $a, b$, and have been a bit lousy to fix that. Anyway, to check my formula i invented the random cubic, $2x^3-5x - 6$, just because it had the root $2$. Now I plug in the values, $b=\frac{5}{2}$ and $a=3$ into my formula, which gives:

$$(\frac{9\sqrt{6} - 19}{6\sqrt{6}})^\frac{1}{3} + (\frac{9\sqrt{6} + 19}{6\sqrt{6}})^\frac{1}{3}$$

After that I was not able to do a single manipulation, apart from noticing that the common denominator is $\sqrt{6}$, however Wolfram Alpha tells me that the value is just simply $2$. Which is my first question, how should I simplify the above expression such that it quickly gives the simplified answer. Infact, I would like a general sure-fire technique, since I encounter this kind of dilemma all the time.

Anyways, I did some Googling thereafter, and came to this: http://www.sosmath.com/algebra/factor/fac11/fac11.html , which showed me how to compute the solution for any kind of cubic. I very much liked the method, though I have some doubts on the substitution, $x=y-\frac{b}{3a}$. Not that its incorrect, I would like to know the motivation from which we obtain that substitution. I would like to know how do we find such beautiful results, which we can use for example to clear the cubic term of a quartic equation.

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    $\begingroup$ There are at least two motivations for letting $x=y-b/3a,$ an algebraic one (substitute $x=y+m$ and ask yourself what should $m$ be so you get a nice cancelling) and a geometric one (making the roots' arithmetic mean zero). $\endgroup$ – Ian Mateus Nov 4 '13 at 18:05
  • $\begingroup$ Oh thanks, Ian, thanks clears up my one point. I would prefer the algebraic one, which when expanded fully gives one term as $y^2(3am+ b)$, so to clear that term I would want $3am + b =0$, which gives me the answer. Wow, so nice! BTW, sorry for slow response my internet connection is very bleak right now. I am going to sleep, will check tomorrow :) $\endgroup$ – Sawarnik Nov 4 '13 at 18:23
  • $\begingroup$ Anyone for the first question? $\endgroup$ – Sawarnik Nov 5 '13 at 11:06

Better late than never. This is strongly related to this answer of mine, when I proved a general form for all solutions of $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n$$ for rational numbers. Here, we should work backwards: we have $$N=\left(\frac{9\sqrt{6} - 19}{6\sqrt{6}}\right)^{1/3} + \left(\frac{9\sqrt{6} + 19}{6\sqrt{6}}\right)^{1/3}=\left(\frac{3}{2}-\frac{19\sqrt 6}{36}\right)^{1/3} + \left(\frac{3}{2}+\frac{ 19\sqrt{6}}{36}\right)^{1/3},$$ so we have $r=6$, $p=3/2$, $q=19/36$. Putting this in the solutions yields $$(p,q)=(3/2, 19/36)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)=\left(\frac{18t^2N+N^3}{8},\,\frac{3N^2t+6t^3}{8}\right)$$ so we have the system $$\begin{cases} 18t^2N+N^3=12 \\ 27N^2t+54t^3=38 \end{cases} $$ with a real solution which can be worked out (nontrivial!) as $(N, t)=(2, 1/3).$ However, I wonder whether solving this system requires essentially the original problem.

  • $\begingroup$ Good, but I don't think calculating the whole stuff would not be daunting. $\endgroup$ – Sawarnik Jan 3 '14 at 13:58
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    $\begingroup$ @Sawarnik I would say that the standard solution would be to find the rational root in the cubic using the rational root test and, then, show that it is the only real root, so the nasty radical and the rational root are equal. Indeed, I asked a question on this system and it seems no easier than Cardano's formula. However, now there is an equivalence of the cubic equation and the system, so I hope it can help you somehow in the future. $\endgroup$ – Ian Mateus Jan 3 '14 at 16:21

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