\( a)\dfrac{{3{x^4} - 2{x^3} - 2{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^4} - 2{x^3} - 6{x^2} + 4{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^2}\left( {{x^2} - 2} \right) - 2x\left( {{x^2} - 2} \right) + 4\left( {{x^2} - 2} \right)}}{{{x^2} - 2}}\\ = \dfrac{{\left( {{x^2} - 2} \right)\left( {3{x^2} - 2x + 4} \right)}}{{{x^2} - 2}}\\ = 3{x^2} - 2x + 4 \)
\( b)\dfrac{{2{x^3} - 26x - 24}}{{{x^2} + 4x + 3}}\\ = \dfrac{{2\left( {{x^3} - 13x - 12} \right)}}{{x + 3x + x + 3}}\\ = \dfrac{{2\left( {{x^3} + {x^2} - {x^2} - x - 12x - 12} \right)}}{{x\left( {x + 3} \right) + x + 3}}\\ = \dfrac{{2\left[ {{x^2}\left( {x + 1} \right) - x\left( {x + 1} \right) - 12\left( {x + 1} \right)} \right]}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {x + 1} \right)\left( {{x^2} - x - 12} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {{x^2} + 3x - 4x - 12} \right)}}{{x + 3}}\\ = \dfrac{{2\left[ {x\left( {x + 3} \right) - 4\left( {x + 3} \right)} \right]}}{{x + 3}}\\ = \dfrac{{2\left( {x + 3} \right)\left( {x - 4} \right)}}{{x + 3}}\\ = 2\left( {x - 4} \right)\\ = 2x - 8\)
\( c)\dfrac{{{x^3} - 7x + 6}}{{x + 3}}\\ = \dfrac{{{x^3} - {x^2} + {x^2} - x - 6x + 6}}{{x + 3}}\\ = \dfrac{{{x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) - 6\left( {x - 1} \right)}}{{x + 3}}\\ = \dfrac{{\left( {x - 1} \right)\left( {{x^2} + x - 6} \right)}}{{x + 3}}\\ = \dfrac{{\left( {x - 1} \right)\left( {{x^2} + 3x - 2x - 6} \right)}}{{x + 3}}\\ = \dfrac{{\left( {x - 1} \right)\left[ {x\left( {x + 3} \right) - 2\left( {x + 3} \right)} \right]}}{{x + 3}}\\ = \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x - 2} \right)}}{{x + 3}}\\ = \left( {x - 1} \right)\left( {x - 2} \right)\\ = {x^2} - 2x - x + 2\\ = {x^2} - 3x + 2 \)
