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When considering the formula for the solution to a degree three polynomial equation, I observe that it involves two different square roots inside of two different third roots. From the standpoint of complex analysis, each square root represents two different possible branches, and each third root represents three different possible branches. So in principle the cubic formula appears to give possibly $3\cdot2\cdot3\cdot2=36$ different roots. Of course we know that a degree three polynomial equation really has only three solutions.

Without using this latter fact, working just off of the formula, is there a way to see that the set of all possible choices of the roots found in the cubic equation reduces to just three numbers?

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  • $\begingroup$ Same goes for the quartic formula as well. Of course the quadratic formula has only one square root, so there is no reduction, or "overlapping" which needs to be explained. $\endgroup$ – Trevor J Richards Sep 18 '17 at 20:45
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    $\begingroup$ They don't: the two cube roots are not independent of each other; choosing them without regard for this will generate non-solutions of the original cubic. $\endgroup$ – Lord Shark the Unknown Sep 18 '17 at 20:50
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    $\begingroup$ encyclopediaofmath.org/index.php/Cardano_formula ... The cube roots must be chosen so $ \alpha \beta = -p/3$ $\endgroup$ – Donald Splutterwit Sep 18 '17 at 20:56
  • $\begingroup$ Look at the derivation of that formula: $x^3+px+q=0$, $x=u+v$, that gives (via $(u+v)^3=u^3+v^3+3uv(u+v)$) $u^3+v^3+q=0$, if we assume $3uv+p=0$. That's the dependency @Lord Shark the Unknown mentioned. $\endgroup$ – Professor Vector Sep 18 '17 at 20:57
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    $\begingroup$ $x^3+px+q$ has the $3$ solutions $x= \alpha +\beta$ where \begin{eqnarray*} \alpha =\sqrt[3]{-\frac{q}{2} \color{blue}{+ \sqrt{ \frac{q^2}{4}+\frac{p^3}{27}}}} \\ \beta =\sqrt[3]{-\frac{q}{2} \color{blue}{- \sqrt{ \frac{q^2}{4}+\frac{p^3}{27}}}} \\ \end{eqnarray*} Notice that if you choose to take the negative square root then $\alpha$ and $\beta$ will simply swap places ( $\alpha \longleftrightarrow \beta$) and give the same value for $x$. The other roots are obtained by choosing appropriate pairs of cube roots of unity as factors of $\alpha$ and $\beta$. $\endgroup$ – Donald Splutterwit Sep 19 '17 at 6:34

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