Let $\alpha$ and $\beta$ be the roots of the quadratic equation $a{x}^2+bx+c=0$ and$\Delta =b^2-4ac$. If $\alpha +\beta ;{\alpha }^2+{\beta }^2;{\alpha }^3+{\beta }^3$ are in geometric progression, then

$\alpha +\beta =-\frac{b}{a};{\alpha }^2+{\beta }^2=\frac{b^2-2ac}{a^2};{\alpha }^3+{\beta }^3=\frac{3abc-b^3}{a^3}$are in geometric progression.

$\Rightarrow \frac{{(b^2-2ac)}^2}{a^4}=\frac{b(b^3-3abc)}{a^4}\Rightarrow ac(b^2-4ac)=0\Rightarrow c\mathrm{\Delta }=0$

or

$\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3 are in G.P$

$\begin{array}{r}\therefore {\left({\alpha }^2+{\beta }^2\right)}^2=(\alpha +\beta )\left({\alpha }^3+{\beta }^3\right)\\ \therefore {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2={\alpha }^4+{\beta }^4+{\alpha }^3\beta +\alpha {\beta }^3\\ \therefore {\alpha }^2{\beta }^2-{\alpha }^3\beta -\alpha {\beta }^3+{\alpha }^2{\beta }^2=0\\ \therefore {\alpha }^2\beta (\beta -\alpha )-\alpha {\beta }^2(\beta -\alpha )=0\\ \therefore \left({\alpha }^2\beta -\alpha {\beta }^2\right)(\beta -\alpha )=0\\ \therefore \alpha \beta (\alpha -\beta )(\beta -\alpha )=0\\ \therefore \alpha \beta (\alpha -\beta {)}^2=0\\ \therefore \alpha =0\text{ or }\beta =0\text{ or }\alpha -\beta =0\end{array}$

$$\begin{gathered} \text{ Case (1) }\alpha =0\text{ or }\beta =0\text{, } \\ \Rightarrow x=0\text{ is a solution of given equation } \\ \therefore c=0 \\ \alpha -\beta =0 \\ \Rightarrow \alpha =\beta \\ \therefore \mathrm{\Delta }=b^2-4ac=0 \end{gathered}$$