Answer
$$ - \frac{{2{x^4} + x}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }}$$
Work Step by Step
$$\eqalign{
& \frac{{x\sqrt {{x^3} - 1} - \frac{{3{x^4}}}{{\sqrt {{x^3} - 1} }}}}{{{x^3} - 1}} \cr
& {\text{Multiply the numerator and denominator by }}\sqrt {{x^3} - 1} \cr
& = \frac{{x{{\left( {\sqrt {{x^3} - 1} } \right)}^2} - \frac{{3{x^4}\sqrt {{x^3} - 1} }}{{\sqrt {{x^3} - 1} }}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} \cr
& = \frac{{x\left( {{x^3} - 1} \right) - 3{x^4}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} \cr
& = \frac{{{x^4} - x - 3{x^4}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} \cr
& = \frac{{ - x - 2{x^4}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} \cr
& = - \frac{{2{x^4} + x}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} \cr} $$
