Answer
$$x = 0,{\text{ }}x = \root 3 \of 4 $$
Work Step by Step
$$\eqalign{
& \frac{{4x\sqrt {{x^3} - 1} - \frac{{3{x^4}}}{{\sqrt {{x^3} - 1} }}}}{{{x^3} - 1}} = 0 \cr
& {\text{Multiply the numerator and denominator by }}\sqrt {{x^3} - 1} \cr
& \frac{{4x{{\left( {\sqrt {{x^3} - 1} } \right)}^2} - \frac{{3{x^4}\sqrt {{x^3} - 1} }}{{\sqrt {{x^3} - 1} }}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} = 0 \cr
& {\text{Simplify}} \cr
& \frac{{4x\left( {{x^3} - 1} \right) - 3{x^4}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} = 0 \cr
& \frac{{4{x^4} - 4x - 3{x^4}}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} = 0 \cr
& \frac{{{x^4} - 4x}}{{\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} }} = 0 \cr
& {\text{Cross product}} \cr
& {x^4} - 4x = 0 \cr
& x\left( {{x^3} - 4} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& x = 0,{\text{ }}{x^3} - 4 = 0 \cr
& x = 0,{\text{ }}x = \root 3 \of 4 \cr} $$
