6. If in the real number range there is
$$
x^{3}+p x+q=(x-a)(x-b)(x-c) \text {, }
$$

and $q \neq 0$, then $\frac{a^{3}+b^{3}+c^{3}}{a b c}=$ $\qquad$ .
Obviously, $a$, $b$, $c$ are the roots of the equation $x^{3} + p x + q = 0$.
By Vieta's formulas, we have $a + b + c = 0$.
By formula (2), we get
$$
\frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a b c} + 3 = 3.
$$