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 $\,\alpha,\beta,\gamma\,$ be roots of the equation, then we have $\,\beta^2=\alpha\gamma\,.$
Also, from Vieta’s relation, we get that
$\alpha\beta\gamma=8\implies\beta^3=8\implies\beta=2\;.$
Now $\,\alpha+\gamma=-2-a\,$ and $\,2\alpha+2\gamma+4=b\,,\,$ hence ,
$\begin{cases}
\alpha+\gamma=-2-a\\
\alpha+\gamma=-2+\dfrac b2
\end{cases}\quad\implies b=-2a\,.$
Since $\,\alpha\,$ and $\,\gamma\,$ are both positive or both negative, it results that $\,|\alpha+\gamma|=|\alpha|+|\gamma|\,$ and $\,|\alpha||\gamma|=\alpha\gamma\,.$
Moreover, by applying the AM-GM inequality, it follows that
$|-2-a|=|\alpha+\gamma|=|\alpha|+|\gamma|\geqslant2\sqrt{|\alpha||\gamma|}=2\sqrt{\alpha\gamma}=4\;\;,$
$|2+a|\geqslant4\;\;,$
$2+a\geqslant4\;\;\;\lor\;\;\;2+a\leqslant-4\;\;,$
$a\geqslant2\;\;\lor\;\;a\leqslant-6\;.$
Consequently the solutions are
$\begin{cases}a\leqslant-6\;\;\lor\;\;a\geqslant2\\b=-2a\quad\qquad\qquad.\end{cases}$
