Answer
$6x^{2}+2x-3-\displaystyle \frac{2}{x}$
Work Step by Step
Using the property: $\displaystyle \quad\frac{A+B}{C}=\frac{A}{C}+\frac{B}{C},$
$\displaystyle \frac{18x^{3}+6x^{2}-9x-6}{3x}=\frac{18x^{3}}{3x}+\frac{6x^{2}}{3x}-\frac{9x}{3x}-\frac{6}{3x}$
$= 6x^{3-1}+2x^{2-1}-3x^{1-1}-2x^{-1}$
= $6x^{2}+2x-3-\displaystyle \frac{2}{x}$
