$\Big(4x-\dfrac{1}{x^3}\Big)^4$
$=\sum\limits_{k=0}^4.C_4^k.(4x)^{4-k}.(-1)^k.\dfrac{1}{x^{3k}}$
$=C_4^0.(4x)^4+C_4^1.(4x)^3.\dfrac{-1}{x^3}+C_4^2.(4x)^2.(-1)^2.\dfrac{1}{x^6}+C_4^3.4x.(-1)^3.\dfrac{1}{x^9}+C_4^4.\dfrac{(-1)^4}{x^{12}}$
$=256x^4-256+\dfrac{96}{x^4}-\dfrac{16}{x^8}+\dfrac{1}{x^{12}}$