Let f(x)=ax^3+bx^2+cx+d, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)

Let $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)
So for the arithmetic progression I took let $\alpha ={x}_{2}$ and r be the ratio of the arithmetic progression.
We have:
${x}_{1}=\alpha -2r,\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha +2r$
Therefore:
${x}_{1}+{x}_{2}+{x}_{3}=-\frac{b}{a}=3\alpha$
${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=9{\alpha }^{2}-2\frac{c}{a}\to 4{r}^{2}=\frac{{b}^{2}-3ac}{3{a}^{2}}$
${x}_{1}{x}_{2}{x}_{3}=\alpha \left({\alpha }^{2}-4{r}^{2}\right)=-\frac{d}{a}$
and we get the final result $2{b}^{3}+27{a}^{2}d-9abc=0$.
How should I take the ratio at the geometric progression for roots?
I tried something like
${x}_{1}=\frac{\alpha }{q},\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha q$
To get ${x}_{1}{x}_{2}{x}_{3}={\alpha }^{3}$ but it doesn't really work out..
Note:
I have to choose from this set of answers:

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erlentzed
Step 1
$a{x}^{3}+b{x}^{2}+cx+d=a\left(x-\frac{\alpha }{q}\right)\left(x-\alpha \right)\left(x-\alpha q\right)$
Expand the rhs to get after simplifications
$a{x}^{3}-\frac{a\alpha \left({q}^{2}+q+1\right)}{q}{x}^{2}+\frac{a{\alpha }^{2}\left({q}^{2}+q+1\right)}{q}x-a{\alpha }^{3}$
Step 2
Compare the coefficients to get
$b=-\frac{a\alpha \left({q}^{2}+q+1\right)}{q}$
$c=\frac{a{\alpha }^{2}\left({q}^{2}+q+1\right)}{q}$
$d=-a{\alpha }^{3}$
Did you like this example?
Widersinnby7
Step 1
To sum up, a cubic has its roots in arithmetic progression if and only if the arithmetic mean of the roots is a root of the cubic ( so equals one of the roots).
For the geometric progression, we could use the same trick, and say that the geometric mean of ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ is a root of P. Alternatively, to avoid cubic roots, one considers the equivalent statement that ${x}_{1}{x}_{2}{x}_{3}$ must equal one of the ${x}_{i}^{3}$. So we set up the equation $Q=0$ with roots ${x}_{1}^{3}$, ${x}_{2}^{3}$, ${x}_{3}^{3}$ and impose the condition that ${x}_{1}{x}_{2}{x}_{3}$ is a root.
The equation for ${x}_{i}^{3}$ can be obtained readily by eliminating x from the equalities $y={x}^{3}$, $a{x}^{3}+b{x}^{2}+cx+d$. We get a cubic equation for y
${a}^{3}{y}^{3}+\left(3{a}^{2}d-3abc+{b}^{3}\right){y}^{2}+\left(3a{d}^{2}-3bcd+{c}^{3}\right)y+{d}^{3}=0$
and the condition is that $-\frac{d}{a}$ is a root of this equation.
Step 2
In general, given a polynomial P of degree n, one can get the condition on the coefficients so that for some ordering of the roots we have an algebraic condition $F\left({x}_{1},\dots ,{x}_{n}\right)=0$. We take all the possible permutations of $F\left({x}_{\sigma \left(1\right)},\dots ,{x}_{\sigma \left(n\right)}\right)$ of F. The condition is that the product of all these permutations is 0.
If we want m conditions ${F}_{1}=\cdots ={F}_{m}=0$, we set up $F=\sum {t}_{i}{F}_{i}$, where ${t}_{1}$, $\dots {t}_{m}$ are variables, consider all the possible permutations of F. The condition is that the product of all these is 0, as a polynomial in ${t}_{1}$, $\dots {t}_{m}$