# Why not write the solutions of a cubic this way?

For the solution of the cubic equation $x^3 + px + q = 0$ Cardano wrote it as:

$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$

but this is ambiguous because it does not tell you which cube roots to match up. Why don't people write it this way today:

$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\frac{-p}{3\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}}$$

which is unambiguous.

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It is as ambiguous as the previous one. If your problem with the original is that you don't know which of the three complex cubic roots to take (or complex square roots to take), why would you know which ones to take in the new expression? You can still take different "cubic roots" for the second fraction. –  Arturo Magidin Dec 30 '10 at 0:42
@Arturo: The difference is that the two cube root expressions in the second form are identical. Think of it as "Let $k=\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}$; $k-\frac{p}{3k}$." –  Isaac Dec 30 '10 at 0:52
@quanta: in UCSMP Precalculus and Discrete Mathematics, 3rd edition, p553 (in the "exploration" question), the cubic formula is given in a form analogous to what you describe. It is given that way for the reason you describe--in particular, because of the way that most calculators and computer algebra systems define the principal root, the "traditional" way of writing the formula does not always yield correct results when computing blindly with technology. –  Isaac Dec 30 '10 at 0:55
Your second form is essentially what my answer to your other cubic equation question does... –  Ｊ. Ｍ. Dec 30 '10 at 0:56
Addendum note to my comment above: The formula as printed in the first printing run of PDM is actually missing a term, though it should be correct in subsequent printing runs. The correct formula reads: Let $A=\frac{\sqrt[3]{-2p^3+9pq-27r+3\sqrt{3}\sqrt{-p^2q^2+4q^3+4p^3r-18pqr+27r^2}}}‌​{3\sqrt[3]{2}}$ and $B=\frac{-p^2+3q}{9A}$. $x_1=-\frac{p}{3}+A-B$, $x_2=-\frac{p}{3}+\frac{-1-i\sqrt{3}}{2}A-\frac{-1+i\sqrt{3}}{2}B$, and $x_3=-\frac{p}{3}+\frac{-1+i\sqrt{3}}{2}A-\frac{-1-i\sqrt{3}}{2}B$ are the solutions to $x^3+px^2+qx+r=0$. –  Isaac Dec 30 '10 at 1:08
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Let $$A=\frac{\sqrt[3]{-2p^3+9pq-27r+3\sqrt{3}\sqrt{-p^2q^2+4q^3+4p^3r-18pqr+27r^2}}}{3\sqrt[3]{2}}$$ and $$B=\frac{-p^2+3q}{9A}.$$ Then, $$x_1=-\frac{p}{3}+A-B,$$ $$x_2=-\frac{p}{3}+\frac{-1-i\sqrt{3}}{2}A-\frac{-1+i\sqrt{3}}{2}B,$$ and $$x_3=-\frac{p}{3}+\frac{-1+i\sqrt{3}}{2}A-\frac{-1-i\sqrt{3}}{2}B$$ are the solutions to $$x^3+px^2+qx+r=0.$$