Answer
$x^{8}+4x^3+\dfrac{6}{x^2}+\dfrac{4}{x^7}+\dfrac{1}{x^{12}}$
Work Step by Step
Apply Binomial Theorem or Binomial expansion for $(x^2+x^{-3})^{4}$.
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
$(x^2+x^{-3})^{4}=\displaystyle \binom{4}{0}(x^2)^{4}(x^{-3})^0+\displaystyle \binom{4}{1}(x^2)^{3}(x^{-3})^1\\+\displaystyle \binom{4}{2}(x^2)^{2}(x^{-3})^2+\displaystyle \binom{4}{3}(x^2)^{1}(x^{-3})^3+\displaystyle \binom{4}{4}(x^2)^{0}(x^{-3})^4$
Thus, $(x^2+x^{-3})^{4}=x^{8}+4x^3+\dfrac{6}{x^2}+\dfrac{4}{x^7}+\dfrac{1}{x^{12}}$
