3. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c=0$ form a geometric sequence, then the value of $a^{3} c-b^{3}$ is
$\qquad$ .
Let the three real roots of the equation be $\frac{k}{q}, k, k q$. By Vieta's formulas, we have:
$$
\left\{\begin{array}{l}
-a=\frac{k}{q}+k+k q=k\left(\frac{1}{q}+1+q\right), \\
b=\frac{k}{q} \cdot k+\frac{k}{q} \cdot k q+k \cdot k q=k^{2}\left(\frac{1}{q}+1+q\right), \\
-c=\frac{k}{q} \cdot k \cdot k q=k^{3}
\end{array}\right.
$$

Therefore, $a^{3} c-b^{3}=k^{3}\left(\frac{1}{q}+1+q\right)^{3} \cdot k^{3}-k^{6}\left(\frac{1}{q}+1+q\right)^{3}=0$.