$$f(x)=\frac{x^2}{(x^3-4)^{\frac13}}$$
Find $a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2$ such that :
\begin{align} & f(x)=a_1x+b_1+\frac {c_1}x+\frac{d_1}{x^2}+o\left(\frac1{x^2}\right) , & x\to \infty\\ & f(x)=a_2x+b_2+\frac {c_2}x+\frac{d_2}{x^2}+o\left(\frac1{x^2}\right) , & x\to -\infty \end{align}
My attempt:
\begin{align} \frac{x^2}{(x^3-4)^{\frac13}} & = \frac{x^2}{x\left(1-\frac4{x^3}\right)^{\frac13}} \\ & = x\left(1-\frac4{x^3}\right)^{-\frac13} \\ & = x\left(1+\frac{4}{3x^3}+\frac{32}{9x^6}+o\left(\frac1{x^6}\right)\right) \\ & = x+\frac4{3x^2}+o\left(\frac1{x^2}\right) \end{align}
$a_1=a_2=1,b_1=b_2=0,c_1=c_2=0,d_1=d_2=\frac43$. Correct ?