The equation $x^3 + 8x^2 - 4x + c = 6x^2 - 12x$ has three roots, one of which is the sum of the other two. What is $c$?

Answer 1

Combining like terms on the left side, we have \begin{align*}
x^3 + (8-6)x^2 +(12-4)x-c&=0,
\end{align*}which becomes $x^3+2x^2+8x-c=0$. Since $x=0$ is not a root, we can divide the equation by $x$ to obtain \[x^2+2x+8-\frac{c}{x}=0.\]If $r$ is a root of this equation, then $\frac{1}{r}$ is also a root. The sum of any two roots of this equation is $-2$. Since $r+r=-2$ means $r=-1$, the remaining root must be $r=\frac{4}{3}$. The numbers $-1$, $-1$, and $\frac{4}{3}$ are the roots in order, so by Vieta's formulas, \[
-1\times -1\times \frac43=\frac{c}{1}=c.
\]Therefore, we have $c=\boxed{-\frac{4}{3}}$.