# Deriving general cube roots

by AndyCav
Tags: cube, deriving, roots
 P: 7 I've been trying to derive general solutions for cubic roots, i.e. the general solutions of (will Latex just work?) $ax^3+bx^2+cx+d=0$ I do not want to be shown the solutions - but does anyone know what direction to go in to achieve this? I thought I'd found solutions at one point but they relied on one of the solutions (the simplest in form) being $\sqrt{-c/a}$ and that doesn't appear to be true! Andy
 P: 7 $$x_{0}=\sqrt{\frac{-c}{a}}$$ Ha ha, I appear to have found out!
 P: 7 Deriving general cube roots This is what I tried: $$f(x)=ax^3+bx^2+cx+d=0$$ is the general equation to solve, and intersects the y axis a distance d from the origin. So, consider $$f(x-x_0)=g(x)$$ where $$x_0$$ is one solution of $$f(x)=0$$. As we have shifted the original function along the x-axis the new function $$g(x)$$ now passes through the origin. After substituting $$x-x_0$$ into f to get g, expand g to get it in the form $$g(x)=px^3+qx^2+rx+s$$. As said above g passes through the origin so that s=0. Therefore, $$s=0=d-ax_0^3+bx_0^2-cx_0$$. Rearranging for d in terms of $$x_0$$ and a, b and c and substituting back into the original form for f we get $$f(x)=a(x^3-x_0^3)+b(x^2-x_0^2)+c(x-x_0)$$. Now, as $$x_0$$ is by definition a solution of f; $$f(x_0)=0$$ we get: $$ax_0^3+cx_0=0$$ which gives the trivial solution $$x_0=0$$ (i.e. d=0) and the two solutions $$x_0=+-\sqrt{\frac{-c}{a}}$$. These don't seem to work. Am I doing something stupid? Andy
 P: 7 There's meant to be a sqaure root sign over that -c/a but it's not coming out on my screen. $$x_0$$ is of course just one solution of the cubic.
 Sci Advisor HW Helper P: 3,684 If you have one solution $$x_0$$ to the cubic, you can just divide by $$x-x_0$$ to reduce it to a quadratic and you're done. If you really want a general solution like Cardanno's, I think the first step would be to remove the $$bx^2$$ term by a change of variable.
 P: 7 Well I'm trying to derive all solutions generally so I don't have $$x_0$$ to begin with. That's a fantastic idea about the change of variable to remove the even term though....! Cheers! I'll have dinner then have another try.