By the distributive property of multiplication,
[tex]\dfrac{9x^3-6x^2+3x}{3x}=\dfrac{9x^3}{3x}-\dfrac{6x^2}{3x}+\dfrac{3x}{3x}[/tex]
(you're multiplying [tex]9x^3-6x^2+3x[/tex] by [tex]\dfrac1{3x}[/tex])
Then cancel common terms that appear in the numerator and denominator - if [tex]a\neq0[/tex], then [tex]\dfrac aa=1[/tex].
[tex]\dfrac{9x^3}{3x}-\dfrac{6x^2}{3x}+\dfrac{3x}{3x}=\dfrac{3x^2\cdot3x}{3x}-\dfrac{2x\cdot3x}{3x}+\dfrac{3x}{3x}=3x^2-2x+1[/tex]
