Originally Posted by

**Quick** So, using your method I get the *incredibly* long equation of (in fact, it's so big I had to set it up horizontally):

...$\displaystyle \small{\sqrt[3]{\frac{9abc-4b^3}{54a^3}+\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\displaystyle \small{+\sqrt[3]{\frac{9abc-4b^3}{54a^3}-\sqrt{\frac{9a^2c^2-6b^2ac+b^4}{81a^4}-\left(\frac{9abc-4b^3}{54a^3}\right)^2}}}$

$\displaystyle \small{-\frac{b}{3a}}$

__.................................................. ................................__

...$\displaystyle x$

Am I right? (or have you never fully written it out before)

Also, clarification, do you mean that this equation always and only works if $\displaystyle \frac{3ac-b^2}{3a^2}\geq0$ or do you mean that it only and works most of the time when $\displaystyle \frac{3ac-b^2}{3a^2}\geq0$?

note: My calculations say $\displaystyle A=\frac{3ac-b^2}{3a^2}$