Answer
$$\frac{{{\text{3}}x{{\left( {{x^2} + 1} \right)}^2}\left( {4 - {x^3} - 3x} \right)}}{{{{\left( {{x^3} + 2} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& \frac{{6x{{\left( {{x^2} + 1} \right)}^2}{{\left( {{x^3} + 2} \right)}^3} - 9{x^2}{{\left( {{x^2} + 1} \right)}^3}{{\left( {{x^3} + 2} \right)}^2}}}{{{{\left( {{x^3} + 2} \right)}^6}}} \cr
& {\text{Divide the numerator and denominator by }}{\left( {{x^3} + 2} \right)^2} \cr
& = \frac{{\frac{{6x{{\left( {{x^2} + 1} \right)}^2}{{\left( {{x^3} + 2} \right)}^3}}}{{{{\left( {{x^3} + 2} \right)}^2}}} - \frac{{9{x^2}{{\left( {{x^2} + 1} \right)}^3}{{\left( {{x^3} + 2} \right)}^2}}}{{{{\left( {{x^3} + 2} \right)}^2}}}}}{{\frac{{{{\left( {{x^3} + 2} \right)}^6}}}{{{{\left( {{x^3} + 2} \right)}^2}}}}} \cr
& = \frac{{6x{{\left( {{x^2} + 1} \right)}^2}\left( {{x^3} + 2} \right) - 9{x^2}{{\left( {{x^2} + 1} \right)}^3}}}{{{{\left( {{x^3} + 2} \right)}^4}}} \cr
& {\text{Factor the numerator, the LCD is 3}}x{\left( {{x^2} + 1} \right)^2} \cr
& = \frac{{{\text{3}}x{{\left( {{x^2} + 1} \right)}^2}\left[ {2\left( {{x^3} + 2} \right) - 3x\left( {{x^2} + 1} \right)} \right]}}{{{{\left( {{x^3} + 2} \right)}^4}}} \cr
& {\text{Simplify}} \cr
& = \frac{{{\text{3}}x{{\left( {{x^2} + 1} \right)}^2}\left( {2{x^3} + 4 - 3{x^3} - 3x} \right)}}{{{{\left( {{x^3} + 2} \right)}^4}}} \cr
& = \frac{{{\text{3}}x{{\left( {{x^2} + 1} \right)}^2}\left( {4 - {x^3} - 3x} \right)}}{{{{\left( {{x^3} + 2} \right)}^4}}} \cr} $$
