6. Given that the equation $x^{3}+a x^{2}+b x+c=$ 0 has three non-zero real roots forming a geometric sequence. Then $a^{3} c-b^{3}=$ $\qquad$
6. 0 .

Solution: Let the three roots be $d, d q, d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b \\
d^{3} q^{3}=-c
\end{array}\right.
$$
(2) ÷ (1) gives $d q=-\frac{b}{a}$, substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$, hence $a^{3} c-b^{3}=0$.