User:MathMan64/TrigConstants
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[edit] Solving Cubic Equations
[edit] 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 
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](../../../../math/f/1/f/f1f64c4eb878cdbf433d2d4a0d40959b.png)
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](../../../../math/7/9/9/799a48d381e68cb2f434dc32d454d1fd.png)
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}}}](../../../../math/2/9/e/29eba8c1859acf0fee762e2ad835e954.png)
Starting with:
![- \frac S {\sqrt[3]{R + \sqrt{R^2+S^3}}}](../../../../math/e/4/9/e494528bf1c4a4707bb402202f4d93ae.png)
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}}}](../../../../math/9/9/4/9944c0b53e52b8a38131601eda0be7a7.png)
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)}}](../../../../math/1/8/8/188136e234dabc4eb5dcd6a5c4f778ed.png)
Add like terms in the denominator:
![\frac {-S \sqrt[3]{R - \sqrt{R^2+S^3}}}{\sqrt[3]{-S^3}}](../../../../math/9/9/4/994deee01c0bbca14ad2026e07c24c31.png)
Finally cancel like factors in the numerator and the demoninator:
![\sqrt[3]{R - \sqrt{R^2+S^3}}](../../../../math/a/0/d/a0d1a72c85192afc0b096502b625f6d9.png)
So the two methods yield the identical results.
