## Precalculus (6th Edition) Blitzer

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