Say I have a sextic equation, but I'm able to get it into the form: $$ax^6 + dx^3 + g = 0$$
I know that I can do a simple substitution like $y = x^3$ to get an equation that I can solve with the Quadratic equation: $$ay^2 + dy + g = 0$$
From there I can get both of $y$'s roots:
- $y_1 = \frac{-d + \sqrt{d^2 - 4ag}}{2a}$
- $y_2 = \frac{-d - \sqrt{d^2 - 4ag}}{2a}$
So, using $y = x^3$, this would mean that two of $x$'s roots are:
- $x_1 = \sqrt[3]{y_1}$
- $x_2 = \sqrt[3]{y_2}$
But there should be at least one more root here according to: https://en.wikipedia.org/wiki/Cubic_function#Vieta.27s_substitution
The quadratic formula allows this to be solved for $w_3$. If $w_1$, $w_2$ and $w_3$ are the three cube roots of one of the solutions in $w_3$...
Can someone help my find at least one more root?
In the complex plane, the equation $z^3=a$ (with $a$ real) has three roots. If as a real number $a=b^3$ (with $b$ a real), then $z^3-b^3=(z-b)(z^2+bz+b^2),$ and then the quadratic equation can be applied to the second factor.