# Problems while solving the cubics

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.

• 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). – Ian Mateus Nov 4 '13 at 18:05
• 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 :) – Sawarnik Nov 4 '13 at 18:23
• Anyone for the first question? – 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.