## Precalculus (6th Edition)

$\color{blue}{r^2-2+\dfrac{1}{r}}$
RECALL: $(a-b)^2=a^2-2ab+b^2$ Use the formula above ith $a=r^{1/2}$ and $b=r^{-1/2}$ to obtain : $=(r^{1/2})^2 -2(r^{1/2})(r^{-1/2}) + (r^{-1/2})^2$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $=(r^{1/2})^2 -2r^{1/2+(-1/2)} + (r^{-1/2})^2 \\=(r^{1/2})^2 -2r^{0} + (r^{-1/2})^2$ Use the rule $a^0=1, a\ne0$ to obtain: $=(r^{1/2})^2 -2(1) + (r^{-1/2})^2 \\=(r^{1/2})^2 -2 + (r^{-1/2})^2$ Use the rule $(a^m)^n=a^{mn}$ to obtain: $=r^{(1/2)(2)} - 2 + r^{(-1/2)(2)} \\=r^1 -2 + r^{-1} \\=r-2 + r^{-1}$ Use the rule $a^{-m} = \dfrac{1}{a^m}$ to obtain: $=r-2+\dfrac{1}{r^1} \\=\color{blue}{r^2-2+\dfrac{1}{r}}$