User:MathMan64/TrigConstants

[edit] Solving Cubic Equations

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.