multiply please!

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\((15x^2-9x-\dfrac{6}{3x^2}-3x-6)(6x^2-6x-\dfrac{12}{2x^2}+2x-4)\\ =(-(4x)-4)(13x^2-(12x)-6)\)

Hope that is correct......

Simplify the following:
(15 x^2-(6 x^2)/3-3 x-9 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

The gcd of 6 and 3 is 3, so 6/3 = (3×2)/(3×1) = 3/3×2 = 2:
(15 x^2-2 x^2-3 x-9 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

Grouping like terms, 15 x^2-2 x^2-3 x-9 x-6 = (15 x^2-2 x^2)+(-9 x-3 x)-6:
(15 x^2-2 x^2)+(-9 x-3 x)-6 (6 x^2-(12 x^2)/2+2 x-6 x-4)

15 x^2-2 x^2 = 13 x^2:
(13 x^2+(-9 x-3 x)-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

-9 x-3 x = -12 x:
(13 x^2+-12 x-6) (6 x^2-(12 x^2)/2+2 x-6 x-4)

The gcd of 12 and 2 is 2, so 12/2 = (2×6)/(2×1) = 2/2×6 = 6:
(13 x^2-12 x-6) (6 x^2-6 x^2+2 x-6 x-4)

Grouping like terms, 6 x^2-6 x^2+2 x-6 x-4 = (-6 x+2 x)-4+(6 x^2-6 x^2):
(-6 x+2 x)-4+(6 x^2-6 x^2) (13 x^2-12 x-6)

2 x-6 x = -4 x:
(13 x^2-12 x-6) (-4 x-4+(6 x^2-6 x^2))

6 x^2-6 x^2 = 0:
-4 x-4 (13 x^2-12 x-6)

Factor -4 out of -4 x-4:
Answer: |-4 (x+1) (13x^2 - 12x - 6)

I think it is meant.

Yes, I meant it like that Omi67. Thank you. Can we do it by factoring out a 3 in the first term instead of working with those annoying fractions?