User:MathMan64/TrigConstants

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.