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 $x^3+a{x}^2+bx+c=0$

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

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

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

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

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.