6. If the polynomial in $x$
$$
y=x^{3}+3 x^{2}+6 x+c
$$

has three roots that can form a geometric progression, then the constant $c=$ $\qquad$
6. 8 .

Let the three roots form a geometric sequence $\alpha, \beta, \gamma$, i.e., $\alpha \gamma=\beta^{2}$ and $\alpha \beta \gamma \neq 0$.
According to Vieta's formulas, we have
$$
\begin{array}{l}
\alpha+\beta+\gamma=-3, \alpha \beta+\beta \gamma+\gamma \alpha=6 \text {. } \\
\text { Hence } 6=\alpha \beta+\beta \gamma+\beta^{2} \\
=\beta(\alpha+\gamma+\beta)=-3 \beta \\
\Rightarrow \beta=-2 \\
\Rightarrow c=-\alpha \beta \gamma=-\beta^{3}=8 .
\end{array}
$$