# Solving a Binomial Sextic

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?

• There are three cubic roots of $y_1$ and also three cubic roots of $y_2$. But all of them can be complex numbers. – kmitov Nov 19 '15 at 13:45
• When you find $x_1$ and $x_2$, there are three cube roots (each multiplied by a (complex) cube root of $1$). This will give you 4 more roots. In general, using Descartes' rule of signs, you will find that the sextic written above has at most $2$ real roots. – Michael Burr Nov 19 '15 at 13:46
• Who said that $y_1$ and $y_2$ are real numbers? They can be complex too. – kmitov Nov 19 '15 at 14:25

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.
• Gah! What is $b$ if $z^3 = a$ and $b^3 = a$ doesn't this necessitate that $b^3 = z^3$ and therefore: $b = z$? – Jonathan Mee Nov 19 '15 at 13:52
• @JonathanMee if $a=\exp(2\pi i/3)$, then $a^2\ne a\ne 1$, $a^2\ne 1$, yet $(a^2)^3=a^3=1^3$. – TZakrevskiy Nov 19 '15 at 14:07
• @TZakrevskiy I'm not sure what $a=exp(\frac{2\pi i}{3})$ is supposed to mean, but it seems like you're trying to communicate that $b^3$ and $z^3$ are not the same. – Jonathan Mee Nov 19 '15 at 14:12
• @JonathanMee No, this is the opposite. When you have $x+iy$ - a complex number, then $\exp(x+iy)=e^x(\cos y+i\sin y)$ by Euler's formula; in my example it gives $a = -\frac 12 + i\frac{\sqrt 3}{2}$. For this complex number $a$ it is easy to check that $a^3=1^3$, however, evidently $a\ne 1$. – TZakrevskiy Nov 19 '15 at 14:35
• @JonathanMee From $b^3=z^3$ IF $b,z$ are assumed to be real numbes then it DOES follow that $b=z.$ However if $b,z$ can be any complex numbers this is not so. – coffeemath Nov 19 '15 at 14:37