I would like to know if there is any way I can get rid of these cubic radicals bellow (1). I am allowing both complex and real values.
$$ \sqrt[3]{ -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } } + \sqrt[3]{ -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } } = x + C\,\,\,\, (1)$$
If it were a second order radical, it wouldn't be as harsh as it is for such radicals above. Since we could raise both sides to $2$.
$$\sqrt{\frac{q^2}{4} + \frac{p^3}{27} } = x+C\,\,\,\, (2)$$
I tried to do the same for (1), however the radicals did not vanish. I also tried this:
$$\sqrt[3]{ -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } } = x+C-\sqrt[3]{ -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } }$$
And then raise to the cube. But no pleasant results.
Hint : For $$A =\sqrt[3]{ -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27} } }$$ your equation can be rewritten as
$$A - \frac{p}{3A} = x+ C$$