<s>[INST] Let $f$ be a real polynomial function find relation between the coefficients such that its roots are in an progression 
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)

So for the arithmetic progression I took let $\alpha = x_2$ and $r$ be the ratio of the arithmetic progression.
We have:
$$x_1=\alpha-2r, \quad x_2=\alpha, \quad x_3=\alpha +2r$$
Therefore:
$$x_1+x_2+x_3=-\frac ba=3\alpha$$
$$x_1^2+x_2^2+x_3^2 = 9\alpha^2-2\frac ca \to 4r^2=\frac {b^2-3ac}{3a^2}$$
$$x_1x_2x_3=\alpha(\alpha^2-4r^2)=-\frac da$$
and we get the final result $2b^3+27a^2d-9abc=0$.
How should I take the ratio at the geometric progression for roots?
I tried something like 
$$x_1=\frac {\alpha}q, \quad x_2=\alpha, \quad x_3=\alpha q$$
To get $x_1x_2x_3=\alpha^3$ but it doesn't really work out..
Note:
I have to choose from this set of answers:
$$\text{(a)} \ a^2b=c^2d \quad\text{(b)}\  a^2b^2=c^2d  \quad\text{(c)}\ ab^3=c^3d$$
$$\text{(d)}\ ac^3=b^3d \quad\text{(e)}\ ac=bd \quad\text{(f)}\ a^3c=b^3d$$
 [/INST] Using your notations
$$ax^3+bx^2+c x+d=a(x-\frac \alpha q)(x-\alpha)(x-\alpha q)$$
Expand the rhs to get after simplifications 
$$a x^3 -\frac{a \alpha  \left(q^2+q+1\right)}{q}x^2+\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}x-a \alpha ^3$$ Compare the coefficients to get
$$b=-\frac{a \alpha  \left(q^2+q+1\right)}{q}$$
$$c=\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}$$
$$d=-a \alpha ^3$$
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