#### Answer

$\color{blue}{r^2-2+\dfrac{1}{r}}$

#### Work Step by Step

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:
$=r-2+\dfrac{1}{r^1}
\\=\color{blue}{r^2-2+\dfrac{1}{r}}$