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