User:MathMan64/TrigConstants

Solving Cubic Equations

Details showing that the two forms are the same

The two plans for solving a cubic equation give what looks like two different solutions. In both plans the formulas for P, Q, R, and S are the same.

Starting from $x^{3}+ax^{2}+bx+c=0\$

$P={\frac {3b-a^{2}}{3}}$

$Q={\frac {9ab-27c-2a^{3}}{27}}$

$R={\frac {Q}{2}}$

$S={\frac {P}{3}}$

Cardano's method result is:

$x={\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}+{\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}-{\frac {a}{3}}$

Vieta's method result is:

$x={\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}-{\frac {S}{\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}}-{\frac {a}{3}}$

These two can be shown to be the same if the second terms of each are identical.

If ${\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}=-{\frac {S}{\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}}$

Starting with:

$-{\frac {S}{\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}}$

Multiply by a unit fraction:

$-{\frac {S}{\sqrt[{3}]{R+{\sqrt {R^{2}+S^{3}}}}}}\cdot {\frac {\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}{\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}}$

Carry through the sum and difference product in the denominators:

${\frac {-S{\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}}{\sqrt[{3}]{R^{2}-(R^{2}+S^{3})}}}$

Add like terms in the denominator:

${\frac {-S{\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}}{\sqrt[{3}]{-S^{3}}}}$

Finally cancel like factors in the numerator and the demoninator:

${\sqrt[{3}]{R-{\sqrt {R^{2}+S^{3}}}}}$

So the two methods yield the identical results.