## Precalculus (6th Edition) Blitzer

$x^2-3x-\frac{1}{x}+3$
Using the Binomial Theorem, we can expand the expression as $(x^{2/3}-\frac{1}{\sqrt[3] x})^3=(x^{2/3}-x^{-1/3})^3=\binom30 (x^{2/3})^3(-x^{-1/3})^0+\binom31 (x^{2/3})^2(-x^{-1/3})^1+\binom32 (x^{2/3})^1(-x^{-1/3})^2+\binom33 (x^{2/3})^0(-x^{-1/3})^3 =x^2-3x+3-x^{-1}=x^2-3x-\frac{1}{x}+3$