Finding a substitution that eliminates the squared term from a cubic equation So I am a little stumped here, and it could be simple. I'm just not exactly sure how to approach this.
The question reads: 

Find $\omega_0$ in the set of Complex numbers, such that the substitution $z = \omega - \omega_0$ reduces the cubic equation $z^3 + Az^2 + Bz +C = 0$ into $\omega^3 -m\omega -n =0$

... where I'm assuming those constants are real numbers.
My first attempt was to just do a straight substitution of $z= \omega - \omega_0$ into the first equation, then expand it, and try to solve that down for what $\omega_0$ should be, but I started to realize that that's probably not the way.
Am I just missing something here?
 A: An alternative to the substitute and expand approach is to use Vieta's formula for the sum of the roots.

Before the substitution, the sum of the roots is $-A$.

After the substitution, the new sum of roots is $-A + 3\omega_0$ (since each root shifts by $\omega_0$).

But since the new quadratic term has zero coefficient, the new sum of roots must be zero, hence
$$-A + 3\omega_0=0 \implies \omega_0 = A/3$$
A: Letting $z=\omega - \omega_0$ and getting
$$ z^3 + Az^2 + Bz + C = \omega^3 + A'\omega^2 +B'\omega +C'$$
Can be accomplished by doing the following synthetic divisions
\begin{array}{r|ccccc}
          & 1
          & A 
          & B
          & C 
\\
-\omega_0 & 0
          & -\omega_0 
          & \omega_0^2 - A\omega_0
          & -\omega_0^3 + A\omega_0^2 - B\omega_0
\\
\hline
          & 1 
          & -\omega_0 + A
          & \omega_0^2 - A\omega_0 + B
          & \color{red}{C'=-\omega_0^3+A\omega_0^2-B\omega_0+C}
\\
-\omega_0 & 0 
          & -\omega_0 
          & 2\omega_0^2 - A\omega_0
\\
\hline
          & 1 
          & -2\omega_0+A 
          & \color{red}{B'=3\omega_0^2-2A\omega_0+B}
\\
-\omega_0 & 0 
          & -\omega_0
\\
\hline
          & 1 
          & \color{red}{A' = -3\omega_0 + A}
\end{array}
So, if we want $A'=0$, then we need $\omega_0 = \frac 13A$
Letting $\omega_0 = \frac 13A$, we get this.
\begin{array}{r|ccccc}
           & 1
           & A 
           & B
           & C 
\\
-\frac 13A & 0
           & -\frac 13A 
           & -\frac 29A^2
           & \frac{2}{27}A^3 - \frac 13AB
\\
\hline
           & 1 
           & \frac 23A
           & -\frac 29A^2 + B
           & \color{red}{C'=\frac{2}{27}A^3-\frac 13AB+C}
\\
-\frac 13A & 0 
           & -\frac 13A 
           & -\frac 19A^2
\\
\hline
           & 1 
           & \frac 13A
           & \color{red}{B'=-\frac 13A^2+B}
\\
-\frac 13A & 0 
           & -\frac 13A
\\
\hline
           & 1 
           & \color{red}{A' = 0}
\end{array}
