Finding a substitution that eliminates the squared term from a cubic equation

Question

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?

The way you suggested will work. Try it. You don't need to expand everything, since you only care about the quadratic term. — quasi, Sep 11 '17 at 03:50
Maybe this can help: https://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic — Ovi, Sep 11 '17 at 03:53
To reduce a real cubic $x^3+Ax^2+Bx+C$ to a depressed cubic you do the substitution $X=x+$ inflection point — Ovi, Sep 11 '17 at 03:54
You know, in looking at the wiki page, I realized that my substitution can can also be in terms of the constants. That never occurred to me. — Ryan Goulden, Sep 11 '17 at 03:55

quasi · Answer 1 · 2017-09-11T04:20:10.193

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 $$-A + 3\omega_0=0 \implies \omega_0 = A/3$$

Steven Alexis Gregory · Accepted Answer · 2017-09-11T15:31:21.793

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}

Finding a substitution that eliminates the squared term from a cubic equation

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