Let $f(x)=a{x}^{3}+b{x}^{2}+cx+d$, 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 $

${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=9{\alpha}^{2}-2\frac{c}{a}\to 4{r}^{2}=\frac{{b}^{2}-3ac}{3{a}^{2}}$

${x}_{1}{x}_{2}{x}_{3}=\alpha ({\alpha}^{2}-4{r}^{2})=-\frac{d}{a}$

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:

$\text{(a)}\text{}{a}^{2}b={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(b)}\text{}{a}^{2}{b}^{2}={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(c)}\text{}a{b}^{3}={c}^{3}d$

$\text{(d)}\text{}a{c}^{3}={b}^{3}d\phantom{\rule{1em}{0ex}}\text{(e)}\text{}ac=bd\phantom{\rule{1em}{0ex}}\text{(f)}\text{}{a}^{3}c={b}^{3}d$

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 $

${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=9{\alpha}^{2}-2\frac{c}{a}\to 4{r}^{2}=\frac{{b}^{2}-3ac}{3{a}^{2}}$

${x}_{1}{x}_{2}{x}_{3}=\alpha ({\alpha}^{2}-4{r}^{2})=-\frac{d}{a}$

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:

$\text{(a)}\text{}{a}^{2}b={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(b)}\text{}{a}^{2}{b}^{2}={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(c)}\text{}a{b}^{3}={c}^{3}d$

$\text{(d)}\text{}a{c}^{3}={b}^{3}d\phantom{\rule{1em}{0ex}}\text{(e)}\text{}ac=bd\phantom{\rule{1em}{0ex}}\text{(f)}\text{}{a}^{3}c={b}^{3}d$