Find all $a,b$ for which the polynomial has real roots and are in geometric progression. Find all $a, b$ such that the roots of $x^3 + ax^2 + bx − 8 = 0$ are real and in a geometric progression.
I did deduce the answer till $a=\dfrac{-b}{2}$.
Using the Vieta's relations I deduced that if $\alpha,\beta$ and $\gamma$ are the roots of the equation (in the same order) then $\beta^2 = \alpha \gamma$, which implies $\beta^3 =8$ or $\beta=2$, and using the other two relations, I found that $a=\dfrac{-b}{2},$ but got stuck while writing the values of $a$ for which the equation satisfies.

Let $\dfrac{2}{r}$, $2$ and $2r$ be the roots.
Since you have $b=-2a$,
$$8r^3+4ar^2-4ar-8=0$$
$$(r-1)[2r^2+(a+2)r+2]=0$$
Now check the discriminant in $2r^2+(a+2)r+2=0$,
$$\Delta=(a+2)^2-16 \ge 0$$
