## Precalculus (6th Edition) Blitzer

$x^{12}+4x^7+6x^2+\dfrac{4}{x^3}+\dfrac{1}{x^8}$
Calculate the Binomial expansion for $(x^3+x^{-2})^{4}$. we have: $(x^3+x^{-2})^{4}=\displaystyle \binom{4}{0}(x^3)^{4}(x^{-2})^0+\displaystyle \binom{4}{1}(x^3)^{3}(x^{-2})^1\\+\displaystyle \binom{4}{2}(x^3)^{2}(x^{-2})^2+\displaystyle \binom{4}{3}(x^3)^{1}(x^{-2})^3+\displaystyle \binom{4}{4}(x^3)^{0}(x^{-2})^4$ or, $=x^{12}+4x^{(9-2)}+6x^{(6-4)}+4x^{(3-6)}+x^{(-8)}$ Hence, we find that $(x^3+x^{-2})^{4}=x^{12}+4x^7+6x^2+\dfrac{4}{x^3}+\dfrac{1}{x^8}$